the current instantaneous growth rate of the culture is 2.5464 bacteria/day.3) The specific rate of change (per capita growth rate) of the culture can be found using the following formula:r = (1/n)(dn/dt) × 100where r is the specific rate of change, n is the number of bacteria per cubic millimeter, dn/dt is the per capita growth rate, and the factor 100 is used to express r as a percentage.
1) The population P of bacteria in the culture is given by the following formula:P = (4/3)πr³n, where r is the radius and n is the number of bacteria per cubic millimeter. Substituting the given values, we get:P = (4/3)π (konst.)³ × 1000 = 4188.79(konst.)³
Hence, the population of bacteria in the culture is 4188.79(konst.)³.2) If the diameter of the culture is increasing at the rate of 2mm per day, then the instantaneous growth rate (in bacteria per day) of the culture can be found as follows:V = (4/3)πr³ is the volume of the culture at time t, and dV/dt is its instantaneous growth rate.
V = (4/3)πr³ = (4/3)π (0.5d)³ = (1/6)πd³
Differentiating both sides with respect to time, we get:
dV/dt = (1/2)πd²(dd/dt)
Substituting the given values, we get:
dd/dt = (2d²/dt)(dV/dt)/(πd⁴)
dd/dt = (2)(2)/(π)(0.5³)
dd/dt = 8/π
dd/dt = 2.5464 bacteria/day
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Let T₂ : P₂ → P₂, be the linear transformation defined by T(P(x))-xp'(x). Find bases for the kernel and cange of the near transformation T.
kernel : {___}
range {___}
State the nulity and rank of T and verify the Rank Theorem.
The linear transformation T₂ : P₂ → P₂ is defined as T₂(P(x)) = xP'(x), where P(x) is a polynomial of degree at most 2. In this problem, we need to find bases for the kernel and range of T₂ and state the nullity and rank of the transformation. Additionally, we need to verify the Rank Theorem.
To find the kernel of T₂, we need to determine the set of polynomials P(x) such that T₂(P(x)) = xP'(x) is the zero polynomial. This means we need to find the polynomials whose derivative is zero, which are constant polynomials. Therefore, the kernel of T₂ consists of all constant polynomials of degree 0. A basis for the kernel is {1}, as any constant polynomial can be represented as a scalar multiple of 1.
To find the range of T₂, we need to determine the set of all polynomials Q(x) that can be obtained as T₂(P(x)) for some polynomial P(x) in the domain. Since T₂(P(x)) = xP'(x), the range of T₂ consists of all polynomials of degree 1. A basis for the range is {x}, as any linear polynomial can be represented as a scalar multiple of x.
The nullity of T₂ is the dimension of the kernel, which is 1 in this case since the kernel has a basis with one element. The rank of T₂ is the dimension of the range, which is also 1 since the range has a basis with one element.
The Rank Theorem states that for a linear transformation from a vector space V to a vector space W, the sum of the nullity (dimension of the kernel) and the rank (dimension of the range) is equal to the dimension of the domain (V). In this case, the dimension of the domain is 3 (degree 2 polynomials), and the sum of the nullity and rank is also 3, satisfying the Rank Theorem.
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In which of the following scenarios is a
dependent t-test used?
Difference in means between two conditions containing different
people, when the data are at least interval and data are normally
dist
In the scenario, "Difference in means between two conditions containing different people, when the data are at least interval and data are normally distributed," a dependent t-test is used.
A dependent t-test is used in the scenario "Difference in means between two conditions containing different people, when the data are at least interval and data are normally distributed."
A dependent t-test is also known as a paired t-test or a repeated-measures t-test. It is a statistical technique that is used to determine whether the mean of the differences between two groups is significant or not. It compares the means of two dependent groups to determine whether there is a significant difference between them.
In the scenario "Difference in means between two conditions containing different people, when the data are at least interval and data are normally distributed," the dependent t-test is used because the two groups contain different people.
The t-test is used to determine whether there is a significant difference between the means of the two groups, which are dependent on each other.
The data in this scenario are at least interval and normally distributed.
Summary:A dependent t-test is used in the scenario "Difference in means between two conditions containing different people, when the data are at least interval and data are normally distributed." It is used to determine whether there is a significant difference between the means of two dependent groups.
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Eig E Mathematics 30-2 6. If y = 7x, x & R, the inverse function is A. y = x7 B. y = logx7 C. y = log7x D. y = log7
The inverse function of y = 7x is y = x/7. None of the options provided, including y = x7, y = logx7, y = log7x, and y = log7, match the correct inverse function.
This means that if we have a function that relates x and y as y = 7x, the inverse function will relate x and y as y = x/7. To find the inverse function, we need to swap the variables x and y in the original equation, y = 7x, resulting in x = 7y. Then, we isolate y by dividing both sides of the equation by 7, giving us y = x/7.
This means that the inverse function of y = 7x is y = x/7. None of the options provided, such as y = x7 (incorrect exponent placement), y = logx7 (logarithm does not match the equation), y = log7x (incorrect logarithm base), or y = log7 (missing variable), represent the correct inverse function for y = 7x.
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A discount of $40 is given off an item marked $70.00 .What change will a customer receive if he or she pays with $100.00
Answer:70.00 is the change
Step-by-step explanation: 40 dollars of 70 is 70-40=30. If the customer pays 100, it would be 100-30=70.
Answer:
To calculate the change that a customer will receive if he or she pays with $100.00 for an item marked $70.00 with a discount of $40, we need to follow these steps:
- First, we need to find the actual price of the item after applying the discount. We can do this by subtracting the discount amount from the original price: $70.00 - $40 = $30.00.
- Next, we need to find the amount of money that the customer pays for the item. Since the customer pays with $100.00, this is simply $100.00.
- Finally, we need to find the difference between the amount paid and the actual price of the item. This is the change that the customer will receive: $100.00 - $30.00 = $70.00.
Therefore, the change that a customer will receive if he or she pays with $100.00 for an item marked $70.00 with a discount of $40 is $70.00.
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Find the functions f (x) and g (a) such that f (g(x)) = (x+6)² - 4.
a) f(x)=x-4.g(x) = (x + 6)² b) g(x) = (x+6)² - 4. f (x) = x c) f(x) = (x+6)² - 4. g(x) = x d) g(x)=x²-4. f(x) = x + 6 e) g(x)=x-4, f(x) = (x + 6)² f) f(x)=x²-4; g(x)=x+6
the only solution is: a) f(x) = x-4, g(x) = (x + 6)²
where f(g(x)) = (g(x)) - 4 = (x + 6)² - 4, which matches the given functional equation.
We can determine the functions f(x) and g(x) by comparing the given functional equation f(g(x)) = (x+6)² - 4 with various forms of the compositions f(g(x)).
a) f(x) = x-4, g(x) = (x + 6)²
f(g(x)) = (g(x)) - 4 = (x + 6)² - 4
This matches the given functional equation, so f(x) = x-4 and g(x) = (x + 6)² is a solution.
b) g(x) = (x+6)² - 4, f(x) = x
f(g(x)) = f((x+6)² - 4) = (x+6)² - 4
This matches the given functional equation, so g(x) = (x+6)² - 4 and f(x) = x is a solution.
c) f(x) = (x+6)² - 4, g(x) = x
f(g(x)) = f(x) = (x+6)² - 4
This does not match the given functional equation, so f(x) = (x+6)² - 4 and g(x) = x is not a solution.
d) g(x) = x²-4, f(x) = x + 6
f(g(x)) =(g(x)) + 6 = (x² - 4) + 6 = x² + 2
This does not match the given functional equation, so g(x) = x² - 4 and f(x) = x + 6 is not a solution.
e) g(x) = x-4, f(x) = (x+6)²
f(g(x)) = f(x-4) = (x-4+6)² = x²
This does not match the given functional equation, so g(x) = x-4 and f(x) = (x+6)² is not a solution.
f) f(x) = x²-4, g(x) = x+6
f(g(x)) = f(x+6) = (x+6)² - 4
This does not match the given functional equation, so f(x) = x²-4 and g(x) = x+6 is not a solution.
Therefore, the only solution is:
a) f(x) = x-4, g(x) = (x + 6)²
where f(g(x)) = (g(x)) - 4 = (x + 6)² - 4, which matches the given functional equation.
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Assume that females have pulse rates that are normally distributed with a mean of u = 74.0 beats per minute and a standard deviation of o=12.5 beats per minute. Complete parts (a) through (c) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 70 beats per minute and 78 beats per minute The probability is 0.2510 (Round to four decimal places as needed.) b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean between 70 beats per minute and 78 beats per minute The probability is a (Round to four decinal places as needed.)
a) The probability that a randomly selected adult female's pulse rate is between 70 and 78 beats per minute is 0.2510.
b) To find the probability that 25 randomly selected adult females have a mean pulse rate between 70 and 78 beats per minute, additional information is needed.
a) To find the probability that a randomly selected adult female's pulse rate is between 70 and 78 beats per minute, we can use the standard normal distribution and calculate the area under the curve between these two values. By converting the values to Z-scores using the formula Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation, we can look up the corresponding area in the Z-table.
Using the given mean (μ = 74.0) and standard deviation (σ = 12.5), we can calculate the Z-scores for 70 and 78 and find the area under the curve between those Z-scores. The resulting probability is 0.2510.
b) To find the probability that 25 randomly selected adult females have a mean pulse rate between 70 and 78 beats per minute, we need additional information, such as the population standard deviation or the distribution of the sample mean. With the provided information, we can only calculate probabilities for individual pulse rates, not for sample means.
To calculate the probability for the mean pulse rate of a sample, we would need the standard deviation of the sample means, also known as the standard error of the mean. Without this information, we cannot determine the probability in part (b).
In summary, the probability that a randomly selected adult female's pulse rate is between 70 and 78 beats per minute is 0.2510. However, without further information, we cannot determine the probability for the mean pulse rate of 25 randomly selected adult females between 70 and 78 beats per minute.
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A ship leaves port on a bearing of 32.0" and travels 12.1 mi. The ship then turns due east and travels 6.6 mi How far is the ship from port, and what is its bearing from port? The ship is mi from the
The distance of the ship from the port is 6.6 miles, and the bearing of the ship from the port is 90°.
Given a ship leaves port on a bearing of 32° and travels 12.1 mi. The ship then turns due east and travels 6.6 mi. The distance of the ship from the port is 6.6 miles
The problem states that, when the ship leaves port it goes on a bearing of 32°. Now, the ship turns due east which means it makes an angle of 90° with the north direction. Thus, we get the final bearing as 90°.Now, we can use sine and cosine functions to calculate the distance of the ship from the port. Let the distance between the ship and port be x.So, sin(90°) = x / 6.6 ⇒ x = 6.6 miand cos(90°) = y / 6.6 ⇒ y = 0 miThus, the ship is 6.6 mi from the port and its bearing from port is 90°.
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What are the x Intercepts for the function? � ( � ) = ( � − 4 ) ( � + 6 ) f(x)=(x−4)(x+6)
Answer:
The x-intercepts are [tex]x=-6[/tex] and [tex]x=4[/tex]
In terms of coordinates, the x-intercepts are (-6,0) and (4,0)
Step-by-step explanation:
The given quadratic function is:
[tex]f(x)=(x-4)(x+6)---(1)[/tex]
To find its x-intercepts, substitute [tex]f(x)=0[/tex] into (1) as follows:
[tex]0=(x-4)(x+6)[/tex]
Then, by the zero-product property, it follows:
[tex]x-4=0= > x=4[/tex]
[tex]x+6=0= > x=-6[/tex]
So, the x-intercepts are [tex]x=-6[/tex] and [tex]x=4[/tex].
In terms of coordinates, the x-intercepts are [tex](-6,0)[/tex] and [tex](4,0)[/tex]
a car sale, cars are selling at the rate of cars per day, where x is the number of days Since the sale began. How many cars will be sold during the first 7 days of the sale? 9. During 12 X+1
During the first 7 days of the sale, the number of cars sold can be calculated by substituting x = 7 into the given equation, resulting in 96 cars.
The rate of car sales is given by the equation f(x) = 12x + 1, where x represents the number of days since the sale began. To find the number of cars sold during the first 7 days of the sale, we need to evaluate the function f(x) for x = 1, 2, 3, 4, 5, 6, and 7 and sum up the values.
For x = 1, f(1) = 12(1) + 1 = 13 cars.
For x = 2, f(2) = 12(2) + 1 = 25 cars.
For x = 3, f(3) = 12(3) + 1 = 37 cars.
For x = 4, f(4) = 12(4) + 1 = 49 cars.
For x = 5, f(5) = 12(5) + 1 = 61 cars.
For x = 6, f(6) = 12(6) + 1 = 73 cars.
For x = 7, f(7) = 12(7) + 1 = 85 cars.
To find the total number of cars sold during the first 7 days, we sum up these values: 13 + 25 + 37 + 49 + 61 + 73 + 85 = 343 cars.
Therefore, during the first 7 days of the sale, 343 cars will be sold.
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A local farmer plants a given number carrots on a certain number of days. We are looking at the number of carrots the farmer can plant over two days. Suppose that the famers must plant at least 4 carrots on the first day, no more than 9 carrots on the second day and farmer has to plant more carrots on the second day than the first day. a) Determine the sample space of the experiment. b) If each of the outcomes in (a) have equal probability of occurring find the probability of the following events: i. Event that there were 13 carrots in total planted over the two days. ii. Event that an odd number of carrots were planted on the second day. c) Are the events (i) and (ii) mutually exclusive? Motivate your answer! d) Are the events (i) and (ii) statistically independent? Motivate your answer! Question 1.2 [2, 2, 21 Suppose that we have two events A and B such that P(4)=0.8 and P(B)=0.7. a) Is it possible that P(AB)=0.1? Explain your answer. b) What is the smallest possible value of P(AB)? c) What is the largest possible value of P(AB)? Question 1.3 [2, 2, 21 Given the following three events A, B and C, find simpler expressions for the following: a) (AUB)(AUB) b) (AUB)(AUB)(AB) c) (AUB)(BUC) Question 1.4 [3.11 A fair coin is tossed three times a) What is the probability of obtaining two or more heads given that there was at least one head is obtained? b) What is the probability of at least one tail? Question 1.5 [4] If B is an event, with P(B)>0, show that the following is true P(AUC|B)=P(A/B)+P(C\B)~P(A^C\B)
Answer:
a) The sample space of the experiment is {(4,5), (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9), (6,7), (6,8), (6,9), (7,8), (7,9), (8,9)}.
b) i. There are 5 outcomes where there are 13 carrots in total planted over the two days: (4,9), (5,8), (6,7), (7,6), (9,4). Therefore, the probability of this event is 5/15 or 1/3.
ii. There are 7 outcomes where an odd number of carrots were planted on the second day: (4,5), (4,7), (5,7), (6,7), (7,5), (7,7), (9,7). Therefore, the probability of this event is 7/15.
c) The events (i) and (ii) are mutually exclusive because there are no outcomes where both events occur.
d) The events (i) and (ii) are not statistically independent because the outcome of event (ii) affects the outcome of event (i). For example, if an odd number of carrots were planted on the second day, it is impossible for there to be an even number of carrots planted over the two days, which is a requirement for event (i) to occur. Therefore, the probability of event (i) is affected by the occurrence of event (ii).
1.2 a) It is not possible that P(AB)=0.1 because the probability of the intersection of two events cannot be greater than the probability of either event occurring alone. In other words, P(AB) ≤ P(A) and P(AB) ≤ P(B).
b) The smallest possible value of P(AB) is 0 because the intersection of two events cannot have a negative probability.
c) The largest possible value of P(AB) is 0.7 because P(AB) cannot be greater than the probability of event B occurring alone.
1.3 a) (AUB)(AUB) = AUB (distributive property)
b) (AUB)(AUB)(AB) = AUB (AB = A∩B, so (AUB)(AUB)(AB) = AUB∩AUB∩B = AUB∩B =
Select the correct answer. What is the expected value per turn for playing Noluz? A. $0.50 B. −$0.17 C. −$0.25 D. −$0.08 E. $0.06
The expected value per turn for playing Noluz is $0.06.
To determine the expected value per turn for playing Noluz, we need to calculate the average outcome (in monetary terms) of each possible outcome and their respective probabilities.
Let's assume that the probabilities and associated outcomes for playing Noluz are as follows:
Outcome 1: Win $1 with probability 0.4
Outcome 2: Lose $0.5 with probability 0.3
Outcome 3: Lose $0.75 with probability 0.2
Outcome 4: Lose $0.25 with probability 0.1
To calculate the expected value, we multiply each outcome by its probability and sum them up:
Expected value = (1 * 0.4) + (-0.5 * 0.3) + (-0.75 * 0.2) + (-0.25 * 0.1)
Expected value = 0.4 - 0.15 - 0.15 - 0.025
Expected value = 0.06
Therefore, the expected value per turn for playing Noluz is $0.06.
The correct answer is E. $0.06.
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Find the price elasticity of demand at the point P=10 for the demand function by the interpretation!
Q = 100 - 3P
The price elasticity of demand measures the responsiveness of the quantity demanded to a change in price. Mathematically, it is defined as the percentage change in quantity demanded divided by the percentage change in price.
In this case, we are interested in finding the price elasticity of demand at the point P = 10. To do this, we need to calculate the percentage change in quantity demanded and the percentage change in price around this point.
Let's start by calculating the percentage change in quantity demanded. The original quantity demanded at P = 10 is given by Q = 100 - 3P, so when P = 10, Q = 100 - 3(10) = 100 - 30 = 70.
Now, let's calculate the new quantity demanded when the price changes slightly. Let's say the new price is P + ΔP, where ΔP represents a small change in price. Using the demand function, the new quantity demanded can be calculated as Q' = 100 - 3(P + ΔP).
The percentage change in quantity demanded can be calculated as (Q' - Q) / Q * 100.
Now, let's calculate the percentage change in price. The original price is P = 10, and the new price is P + ΔP. The percentage change in price can be calculated as (ΔP / P) * 100.
Finally, we can calculate the price elasticity of demand at P = 10 using the formula: Price Elasticity of Demand = (Percentage change in quantity demanded) / (Percentage change in price).
By interpreting the price elasticity of demand at the point P = 10, we can determine the responsiveness of the quantity demanded to a change in price in that specific scenario.
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Find the dimension of a closed rectangular box that has a square
base and capacity of 27in^3. And is constructed with the least
amount of material.
Given that the closed rectangular box has a square base and a capacity of 27 in³ and it is constructed with the least
amount of material. Now, we have to find the dimensions of the box.To find the dimensions of the box we need to use the following formula:V = lwh ...(1)whereV = volume of the rectangular boxl = length of the boxw = width of the boxh = height of the boxGiven that, V = 27 in³ and the base of the box is a square. That is, l = wUsing this in equation (1), we get27 = l²h27 = w²hNow we need to minimize the surface area.
The surface area can be given by the formula:S.A. = 2lw + 2lh + 2whwhere S.A. = Surface Area of the box.Now substituting l = w in equation (1),
we get27 = l²h27 = w²h
Then, h = 27 / l² ...(2)Substituting equations (1) and (2) in surface area, we get:S.A. = 2lw + 2lh + 2wh= 2lw + 2l(27 / l²) + 2w (27 / l²)= 2l²w⁻¹ + 54l⁻¹ + 54w⁻¹Now we need to minimize S.A. with respect to l. That is we need to find dS.A./dlS.A. = 2l²w⁻¹ + 54l⁻¹ + 54w⁻¹Differentiating w.r.t l,dS.A./dl = 4lw⁻¹ - 54l⁻²Now to find the minimum value, we have to equate the derivative to zero.(dS.A./dl) = 4lw⁻¹ - 54l⁻² = 0or4 / l = 54 / w²Multiplying both sides with l² / 4, we getl² / 4 = 54 / w²l = 6w / √3Putting this value of l in equation (1), we get:27 = l²h27 = (6w / √3)²h27 = 12w²h/3h = 9 / w²Now, we need to minimize S.A. with respect to w. That is we need to find dS.A./dwS.A. = 2lw + 2lh + 2wh= 2lw + 2l(9 / w²) + 2ww⁻¹= 2lw + 18w⁻¹ + 2wNow differentiating w.r.t w,dS.A./dw = 2l w⁻¹ - 18w⁻² + 2Differentiating w.r.t w again to find whether it is maximum or minimum, we get:d²S.A./dw² = -2lw⁻² + 36w⁻³The value of d²S.A./dw² is negative. Hence the given equation has a maximum.So, to minimize the surface area, the value of l and w should be equal.
So, l = w = 3√3.Then h = 9 / (3√3)² = 1√3∴ The dimensions of the box are 3√3 x 3√3 x 1√3 cubic inches.
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You work for a nuclear research laboratory that is contemplating leasing a diagnostic scanner (leasing is a very common practice with expensive, high-tech equipment). The scanner costs $4,700,000, and it would be depreciated straight-line to zero over four years. Because of radiation contamination, it actually will be completely valueless in four years. You can borrow at 7 percent before taxes. Your company does not anticipate paying taxes for the next several years, but the leasing company has a tax rate of 22 percent. Over what range of lease payments will the lease be profitable for both parties? (Do not round intermediate calculations and enter your answers from lowest to highest rounded to 2 decimal places, e.g., 32.16.) Total payment range to
The range of lease payments is empty or non-existent in this case.
To determine the range of lease payments that will be profitable for both parties, we need to compare the costs and benefits associated with the lease.
1. Calculate the Depreciation Expense:
The scanner costs $4,700,000 and will be depreciated straight-line to zero over four years. Therefore, the annual depreciation expense is:
Depreciation Expense = Cost of Scanner / Useful Life = $4,700,000 / 4 = $1,175,000 per year.
2. Calculate the Lease Payments:
Let's denote the lease payment as P. The lease payments will be made for four years.
3. Calculate the After-Tax Lease Payments:
Since the leasing company has a tax rate of 22 percent, the after-tax lease payment can be calculated as:
After-Tax Lease Payment = Lease Payment * (1 - Tax Rate) = P * (1 - 0.22) = 0.78P.
4. Calculate the Borrowing Cost:
The company can borrow at an interest rate of 7 percent before taxes.
5. Determine the Profitability Condition:
For the lease to be profitable for both parties, the after-tax lease payments should be less than or equal to the borrowing cost.
0.78P ≤ 0.07P
Solving the inequality, we find:
P ≤ 0
This inequality suggests that there is no range of lease payments that will be profitable for both parties. The lease would not be profitable under the given conditions.
Therefore, the range of lease payments is empty or non-existent in this case.
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It is known that the grade point avarage (GPA) of students among all those graduating from a university in 2020 had the mean of 3.22, and the standard deviation of 0.26.
a. Compute the probability that a randomly selected GPA score from the population is between 2.5 and 3.5.
b. Find the GPA score that is the 82th percentile.
c. Find the interquartile range (IQR) of the GPA. d. For n=100 randomly selected students, find the probability that the sample mean of GPA is between 2.5 and 3.5.
To compute the probability that a randomly selected GPA score from the population is between 2.5 and 3.5, we can use the standard normal distribution which will come out to be 3.43
To find the GPA score that is the 82nd percentile, we need to find the z-score that corresponds to the 82nd percentile. We can use the inverse standard normal distribution or the z-score formula. The z-score corresponding to the 82nd percentile is approximately 0.93. Using the formula z = (x - mean) / standard deviation, we can solve for x, the GPA score. Rearranging the formula, we have x = z * standard deviation + mean. Substituting the values, x = 0.93 * 0.26 + 3.22 = 3.43.
The interquartile range (IQR) is a measure of the spread of a distribution. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1). Since the GPA distribution is not provided, we cannot directly calculate the quartiles. However, if we assume a normal distribution, we can estimate the quartiles using the mean and standard deviation. Q1 would be approximately the mean minus 0.67 times the standard deviation, and Q3 would be approximately the mean plus 0.67 times the standard deviation. The IQR would then be the difference between Q3 and Q1.
To find the probability that the sample mean of GPA is between 2.5 and 3.5 for a sample of 100 students, we can use the Central Limit Theorem. According to the theorem, for sufficiently large sample size, the distribution of sample means approaches a normal distribution, regardless of the shape of the population distribution. Since the sample size is large (n = 100) and the population standard deviation is known, we can calculate the standard error of the mean using the formula standard deviation/sqrt (n). Then, we can standardize the values of 2.5 and 3.5 using the sample mean and the standard error of the mean, and find the probability using a standard normal distribution table or a calculator.
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Let X be a discrete random variable with the following PMF
PX(x)=0.10.20.20.30.20for x=0.2for x=0.4for x=0.5for x=0.8for x=1otherwise
Find RX the range of the random variable X
a. Find P(X≤0.5)
b. Find P(0.25
c. Find P(X=0.2|X<0.6)
The range of a random variable X is the set of all possible values that X can take. In this case, the range is {0, 0.2, 0.4, 0.5, 0.8, 1}.
a. To find P(X ≤ 0.5), we sum up the probabilities of all values less than or equal to 0.5:
P(X ≤ 0.5) = P(X = 0) + P(X = 0.2) + P(X = 0.4) + P(X = 0.5)
= 0.1 + 0.2 + 0.2 + 0.3
= 0.8
b. To find P(0.25 < X < 0.8), we sum up the probabilities of all values between 0.25 and 0.8 (excluding the endpoints):
P(0.25 < X < 0.8) = P(X = 0.4) + P(X = 0.5)
= 0.2 + 0.3
= 0.5
c. To find P(X = 0.2 | X < 0.6), we need to calculate the conditional probability of X = 0.2 given that X is less than 0.6. We first calculate the probability of X being less than 0.6:
P(X < 0.6) = P(X = 0) + P(X = 0.2) + P(X = 0.4) + P(X = 0.5)
= 0.1 + 0.2 + 0.2 + 0.3
= 0.8
Then we calculate the probability of X = 0.2 given X < 0.6:
P(X = 0.2 | X < 0.6) = P(X = 0.2 and X < 0.6) / P(X < 0.6)
= P(X = 0.2) / P(X < 0.6)
= 0.2 / 0.8
= 0.25
Therefore, P(X = 0.2 | X < 0.6) is 0.25.
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Find the general solution of the differential equation: dy/dt=−2ty+4e^−t^2
What is the integrating factor? μ(t)=
Use lower case c for the constant y(t)=
Therefore, the general solution of the differential equation is `y(t) = e^t^2(C + 4Ei(-t^2))` where `C` is the constant.
To find the general solution of the differential equation `dy/dt = −2ty + 4e^−t^2`, we need to find the integrating factor and then multiply the given differential equation by it and integrate both sides.
Using the formula, μ(t) = `e^(∫-2t dt)`= `e^-t^2`The integrating factor is `μ(t) = e^-t^2`.
Multiplying both sides of the given differential equation by the integrating factor yields: `e^-t^2 dy/dt - 2tye^-t^2 = 4`
The left-hand side is the product rule of `(e^-t^2 y(t))'`.
Integrating both sides yields: ∫`(e^-t^2 dy/dt - 2tye^-t^2) dt = ∫ 4 dt `Using the product rule on the left-hand side gives: e^-t^2 y(t) = `∫ 4e^t^2 dt/ e^-t^2` Using integration by substitution, let `u = -t^2`. Then, `du/dt = -2t` and `dt = -du/2t`.
The integral becomes: e^-t^2 y(t) = `∫-4 e^u du/2u` = `-2∫ e^u du/u`
This is the definition of the exponential integral function `Ei(u)`, so:∫e^-t^2 dy/dt - 2tye^-t^2 dt = 4Ei(-t^2) + C, where C is a constant of integration. Dividing by the integrating factor `μ(t)` and simplifying gives: y(t) = `e^t^2(C + 4Ei(-t^2))`
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Given differential equation is,dy/dt = -2ty + 4e^(-t²). The general solution of the given differential equation is y(t) = (4t + C) * e^(-t²).
We can write it as dy/dt + 2ty = 4e^(-t²)
To find the integrating factor (μ(t)), we need to multiply the equation by an integrating factor.I.F. (μ(t)) = e^(∫2t dt)I.F. (μ(t)) = e^(t²)
Multiplying both sides of the differential equation by μ(t)we get, e^(t²)dy/dt + 2tye^(t²) = 4e^(-t²) * e^(t²)
Simplifying the above equation, we get,d/dt [y * e^(t²)] = 4
Then, integrating both sides, we gety * e^(t²) = 4t + C
where C is the constant of integration.
Dividing both sides by e^(t²), we get,y(t) = (4t + C) * e^(-t²)
Where c is the constant of integration.
Therefore, the integrating factor is μ(t) = e^(t²)
The general solution of the given differential equation is y(t) = (4t + C) * e^(-t²).
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QUESTION 5
If the average daily income for small grocery markets in Riyadh
is 5000 riyals, and the standard deviation is 900 riyals, in a
sample of 1600 markets find the standard error of the mean?
The standard error of the mean is 22.5 riyals.
The given information is as follows:
The average daily income for small grocery markets in Riyadh is 5000 riyals.
The standard deviation is 900 riyals.
In a sample of 1600 markets find the standard error of the mean.
To calculate the standard error of the mean, we will use the following formula:
SE = \frac{s}{\sqrt{n}}
where s is the sample standard deviation and n is the sample size.
We have the sample standard deviation s = 900 and the sample size n = 1600.
Putting these values in the formula, we get:
SE = \frac{900}{\sqrt{1600}}
SE = \frac{900}{40} = 22.5
Therefore, the standard error of the mean is 22.5 riyals.
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(1 point) Solve the problem PDE: Utt = 81UIT BC: u(0, t) = u(1, t) = 0 IC: u(x,0) = 8 sin(27x), u(x, t) = help (formulas) 00 u₁(x,0) = 3 sin(3πx)
The solution to the given PDE is \[u(x, t) = 24\sum_{n=1}^\infty \sin 3n\pi x\sin 9n\pi t\].
The given partial differential equation is, \[U_{tt} = 81U_{xx}\]with boundary conditions, \[u(0, t) = u(1, t) = 0\]and initial conditions,\[u(x, 0) = 8 \sin (27x),\;\;u_t(x, 0) = 0.\]The solution to the PDE can be found using the method of separation of variables as follows:Assume that the solution to the PDE can be expressed as a product of two functions, namely\[u(x, t) = X(x)T(t)\]Substituting this into the given PDE, we get,\[XT'' = 81 X''T\]Dividing both sides by XT, we get,\[\frac{T''}{81T} = \frac{X''}{X}\]Let the constant of separation be $-\lambda^2$.Then we can write,\[\begin{aligned} \frac{T''}{81T} &= -\lambda^2\\ T'' + 81\lambda^2T &= 0 \end{aligned}\]The solution to this ODE is,\[T(t) = c_1\cos 9\lambda t + c_2\sin 9\lambda t\]Using the boundary conditions, we can conclude that $c_1 = 0$.
Using the initial condition, we can write,\[\begin{aligned} u(x, 0) &= 8\sin (27x)\\ X(x)T(0) &= 8\sin (27x)\\ AT(0)\sin 3\lambda x &= 8\sin (27x) \end{aligned}\] Comparing coefficients, we get,\[AT(0) = \frac{8}{\sin 3\lambda x}\]Differentiating both sides with respect to time, we get,\[A\frac{d}{dt}(T(t))\sin 3\lambda x = 0\]Using the initial condition for $u_t$, we have,\[u_t(x, 0) = 0 = c_2 9\lambda A \sin 3\lambda x\]Therefore, we must have $\lambda = n$ where $n$ is an integer.We have,\[\begin{aligned} AT(0) &= \frac{8}{\sin 3nx}\\ &= 24\sum_{k=0}^\infty (-1)^k\frac{\sin (6k+3)n\pi x}{(6k+3)n\pi} \end{aligned}\] Hence, we get the solution,\[\begin{aligned} u(x, t) &= \sum_{n=1}^\infty X_n(x)T_n(t)\\ &= 24\sum_{n=1}^\infty \sin 3n\pi x\sin 9n\pi t \end{aligned}\].
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If a man normally consuming 2600 kcals per day reduces his intake to 1500 kcals per day, how much weight will he lose in one week?
a. .5 kg
b. 1.0 kg
c. 1.5 kg
d. 2.0 kg
The man is expected to lose approximately 2.2 kg in one week. None of the provided answer options exactly match this result, so the closest option would be d. 2.0 kg.
To determine the weight loss of a person based on calorie reduction, we need to consider the calorie deficit created by the reduction in daily intake. One pound (0.45 kg) of body weight is roughly equivalent to a calorie deficit of 3500 calories. Therefore, the weight loss can be calculated as follows:
Calorie deficit per day = Initial calorie intake - Reduced calorie intake
Calorie deficit per week = Calorie deficit per day * 7
Weight loss (in kg) = Calorie deficit per week / 3500
Given that the man normally consumes 2600 kcals per day and reduces his intake to 1500 kcals per day, we can calculate the calorie deficit and weight loss:
Calorie deficit per day = 2600 - 1500 = 1100 calories
Calorie deficit per week = 1100 * 7 = 7700 calories
Weight loss = 7700 / 3500 = 2.2 kg (approximately)
Therefore, the man is expected to lose approximately 2.2 kg in one week. None of the provided answer options exactly match this result, so the closest option would be d. 2.0 kg.
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Find all solutions of the equation m ⁿ= nᵐ, where m and n are positive integers (Hint: write m = p₁ᵃ¹... pᵣᵃʳ and n = pi...p where p₁ᵇ¹, ..., pᵣᵇʳ are primes).
Show that if a, b, c ∈ Z with c > 0 such that a = b (mod c), then (a, c) = (b, c).
The solutions to the equation mⁿ = nᵐ, where m and n are positive integers, are m = n or m = n = 1. The equation has no other solutions.
To solve the equation mⁿ = nᵐ, we can consider the prime factorizations of m and n. We can write m = p₁ᵃ¹... pᵣᵃʳ and n = p₁ᵇ¹... pᵣᵇʳ, where p₁, ..., pᵣ are distinct primes.
Since mⁿ = nᵐ, we have (p₁ᵃ¹... pᵣᵃʳ)ⁿ = (p₁ᵇ¹... pᵣᵇʳ)ᵐ. For this equation to hold, the exponents must be equal for each prime factor. Therefore, we have a system of equations:
a₁n = b₁ᵐ
a₂n = b₂ᵐ
...
aᵣn = bᵣᵐ
From these equations, it follows that aᵢ divides bᵢᵐ for each i, and bᵢ divides aᵢn. This implies that aᵢ divides bᵢᵐ and bᵢ divides aᵢn, so aᵢ = bᵢ. Therefore, m = n.
The only other possibility is when m = n = 1. In this case, 1ⁿ = 1ⁿ is always true.
Hence, the solutions to the equation are m = n or m = n = 1, and there are no other solutions.
Regarding the second statement, if a = b (mod c), it means that a and b have the same remainder when divided by c. This implies that c divides both a - b and b - a. Therefore, (a, c) = (b, c) = c, as c is the greatest common divisor of a and c as well as b and c.
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Let X denote the amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function kx, if 0 ≤ x ≤ 1 f(x) = otherwise. a. Find the value of k. Calculate the following probabilities: b. P(X1), P(0.5 ≤ x ≤ 1.5), and P(1.5 ≤ X) [3+5]
Given, X denotes the amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function kx, if 0 ≤ x ≤ 1 f(x) = otherwise.a)
To find the value of k, we use the property of density function that the integral of density function over its range is 1. i.e. ∫ f(x) dx = 1 for all x in [a,b] ∫ kx dx = 1 for all x in
[0,1] ⇒ k/2 [x^2]0¹ = 1 (1/2) [1^2] - (1/2) [0^2] = 1 (1/2) - (0) = 1/2 ∴ k = 2b)
;a. k = 2b. i. P(X1) = 1, ii. P(0.5 ≤ x ≤ 1.5) = 2 and iii. P(1.5 ≤ X) = 0
Hence, X denotes the amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function kx, if 0 ≤ x ≤ 1 f(x) = otherwise.a)
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A 6.00 x 105 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.800 m by a large spring bumper at the end of its track. What is the force constant k of the spring (in N/m)?
To find the force constant k of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed as:
F = -kx
where F is the force exerted by the spring, k is the force constant (also known as the spring constant), and x is the displacement of the spring.
In this scenario, the subway train is brought to a stop by the spring bumper, so the force exerted by the spring is equal to the force required to stop the train. We can use the equation for force to find the force constant.
Given:
Mass of the subway train (m) = 6.00 x 10^5 kg
Initial velocity (v₀) = 0.500 m/s
Displacement (x) = 0.800 m
The force required to stop the train can be calculated using Newton's second law:
F = ma
where F is the force, m is the mass, and a is the acceleration.
In this case, the train is brought to a stop, so its final velocity is zero. The acceleration can be calculated using the kinematic equation:
v² = v₀² + 2ax
Since the final velocity is zero, we can rewrite the equation as:
0 = v₀² + 2ax
Solving for acceleration (a), we have:
a = -v₀² / (2x)
Substituting the given values:
a = -(0.500 m/s)² / (2 * 0.800 m)
a = -0.15625 m/s²
Now, we can calculate the force:
F = ma
F = (6.00 x 10^5 kg) * (-0.15625 m/s²)
F = -9.375 x 10^4 N
According to Hooke's Law, this force is equal to -kx. Comparing the equation with the calculated force:
-9.375 x 10^4 N = -k * 0.800 m
Solving for the force constant (k):
k = (-9.375 x 10^4 N) / (0.800 m)
k = -1.171875 x 10^5 N/m
Therefore, the force constant of the spring is approximately -1.171875 x 10^5 N/m.
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Find the first five terms (ao, a1, A₂, A3, A4) of the fourier series of the function fox)= e^x con the interval [-x, x].
The first five terms of the Fourier series of f(x) = e^x on [-x, x] are: a0 = e^x - e^(-x) a1 = e^x (cos(x) + sin(x)) - e^(-x) (cos(x) - sin(x)) a2 = e^x cos(2x) + 2sin(2x) - cos(2x) + 1 a3 = e^x cos(3x) + 3sin(3x) - cos(3x) + 1 a4 = e^x cos(4x) + 4sin(4x) - cos(4x) + 1
The first five terms of the Fourier series of the function f(x) = e^x on the interval [-x, x] are given by:
a0 = 1/2 ∫[-x,x] e^x dx = 1/2 [e^x] from -x to x = e^x - e^(-x) a1 = 1/2 ∫[-x,x] e^x cos(x) dx = 1/2 [e^x cos(x) + sin(x)] from -x to x = e^x (cos(x) + sin(x)) - e^(-x) (cos(x) - sin(x))a2 = 1/2 ∫[-x,x] e^x cos(2x) dx = 1/2 [2e^x cos(2x) + (4sin(2x) - 2cos(2x))] from -x to x = e^x cos(2x) + 2sin(2x) - cos(2x) + 1a3 = 1/2 ∫[-x,x] e^x cos(3x) dx = 1/2 [3e^x cos(3x) + (9sin(3x) - 3cos(3x))] from -x to x = e^x cos(3x) + 3sin(3x) - cos(3x) + 1a4 = 1/2 ∫[-x,x] e^x cos(4x) dx = 1/2 [4e^x cos(4x) + (16sin(4x) - 4cos(4x))] from -x to x = e^x cos(4x) + 4sin(4x) - cos(4x) + 1
Therefore, the first five terms of the Fourier series of f(x) = e^x on [-x, x] are: a0 = e^x - e^(-x) a1 = e^x (cos(x) + sin(x)) - e^(-x) (cos(x) - sin(x)) a2 = e^x cos(2x) + 2sin(2x) - cos(2x) + 1 a3 = e^x cos(3x) + 3sin(3x) - cos(3x) + 1 a4 = e^x cos(4x) + 4sin(4x) - cos(4x) + 1
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(a) Attendance at the Accra Sports Stadium was alysed by the General Secretary, Prosper Harrison Addo. The analysis demonstrated that spectators consisted of 70% males. If seven people are randomly selected from the spectators during a football match, What is the probability that 4 of them are males? (3 marks) i 11. Find the probability that at most 5 of them are females (4 marks)
a) The probability of randomly selecting 4 males out of 7 spectators, given that 70% of the spectators are males, can be calculated using the binomial probability formula.
b) To find the probability that at most 5 of the randomly selected spectators are females, we need to calculate the cumulative probability of selecting 0, 1, 2, 3, 4, and 5 females from the total number of selected spectators.
a) To calculate the probability of selecting 4 males out of 7 spectators, we can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- n is the total number of trials (number of people selected)
- k is the number of successful trials (number of males selected)
- p is the probability of success in a single trial (probability of selecting a male)
- C(n, k) is the binomial coefficient, calculated as C(n, k) = n! / (k! * (n - k)!)
In this case, n = 7, k = 4, and p = 0.70 (probability of selecting a male). Therefore, the probability of selecting 4 males out of 7 spectators is:
P(X = 4) = C(7, 4) * (0.70)^4 * (1 - 0.70)^(7 - 4)
b) To find the probability that at most 5 of the selected spectators are females, we need to calculate the cumulative probability of selecting 0, 1, 2, 3, 4, and 5 females. This can be done by summing the individual probabilities for each case.
P(X ≤ 5 females) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
To calculate each individual probability, we use the same binomial probability formula as in part a), with p = 0.30 (probability of selecting a female).
Finally, we sum up the probabilities for each case to find the probability that at most 5 of the selected spectators are females.
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The process of making chairs consists of five operations: cleaning, cutting, bonding, painting, and finishing. The standard timings of each operation is: 0.52, 0.48, 0.65, 0.41, and 0.55 minute. The througput yield of each process is 0.99. Assuming a demand of 700 chairs per week with 5 working days with 8 hours/day, a. Will the current process be able to meet the demand? What is the efficiency of the current process? b. If the process can be balanced without reducing any time, can it meet the demand? What would be the balanced standard time? c. What is the sigma level of the process
The efficiency of the current process is 0.00000152
a. To determine if the current process can meet the demand, we need to calculate the total time required to produce 700 chairs per week.
Total time = Demand per week * Total working time per chair
Demand per week = 700 chairs
Total working time per chair = 5 working days * 8 hours/day * 60 minutes/hour
Total time = 700 * (5 * 8 * 60) = 1,680,000 minutes
The total time required for production is 1,680,000 minutes.
Now, we can calculate the total time available for production by considering the throughput yield of each process.
Total time available = Standard time of each operation * Throughput yield of each operation
Standard time of each operation = 0.52 + 0.48 + 0.65 + 0.41 + 0.55 = 2.61 minutes
Total time available = 2.61 * (0.99)^5 = 2.56 minutes
Since the total time required (1,680,000 minutes) is greater than the total time available (2.56 minutes), the current process will not be able to meet the demand.
The efficiency of the current process can be calculated as:
Efficiency = Total time available / Total time required
Efficiency = 2.56 / 1,680,000 ≈ 0.00000152
b. If the process can be balanced without reducing any time, the balanced standard time would be the average of the standard times of each operation.
Balanced standard time = (0.52 + 0.48 + 0.65 + 0.41 + 0.55) / 5 = 0.522 minutes
To determine if the balanced process can meet the demand, we need to calculate the total time available using the balanced standard time:
Total time available = Balanced standard time * (Throughput yield of each operation)^5
Total time available = 0.522 * (0.99)^5 ≈ 0.515 minutes
Since the total time required (1,680,000 minutes) is still greater than the total time available (0.515 minutes), the balanced process will not be able to meet the demand.
c. The sigma level of the process can be calculated using the formula:
Sigma level = (Total time available - Total time required) / (Standard deviation of the process)
To calculate the standard deviation, we need the standard deviation of each operation. If the standard deviations are not provided, we cannot determine the sigma level of the process.
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The number of machine breakdowns per day at Yuwen Chen's factory is either 0, 1, or 2, with probabilities 0.3, 0.3, or 0.4, respectively. The following random numbers have been generated: 35, 41, 81, 76, 44, 17, 3, 29, 89, and 17. (Note: Assume the random number interval begins at 01 and ends at 00.)
Based on the given probabilty distribution, the number of breakdowns for the given random number are: Random Number Number of Breakdowns
35 ___
41 ___
81 ___
76 ___
44 ___
17 ___
3 ___
29 ___
89 ___
17 ___
Proportion of days that had at least one breakdown = ____% (round your response to the nearest whole number).
Therefore, the proportion of days that had at least one breakdown is 40%.
To determine the number of breakdowns corresponding to each random number, we compare the random number with the cumulative probabilities of the given probability distribution.
The cumulative probabilities for the number of breakdowns are as follows:
P(0 breakdowns) = 0.3
P(0 or 1 breakdown) = 0.3 + 0.3 = 0.6
P(0, 1, or 2 breakdowns) = 0.3 + 0.3 + 0.4 = 1.0
Using the given random numbers and the cumulative probabilities, we can determine the number of breakdowns for each random number:
35: Number of breakdowns = 1
41: Number of breakdowns = 1
81: Number of breakdowns = 2
76: Number of breakdowns = 2
44: Number of breakdowns = 1
17: Number of breakdowns = 0
3: Number of breakdowns = 0
29: Number of breakdowns = 0
89: Number of breakdowns = 2
17: Number of breakdowns = 0
To calculate the proportion of days that had at least one breakdown, we count the number of days with one or more breakdowns and divide it by the total number of days (which is equal to the total number of random numbers generated).
Number of days with at least one breakdown = 4 (35, 41, 81, 76)
Total number of days = 10
Proportion of days that had at least one breakdown = (4 / 10) * 100% = 40%
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Consider the following two systems. a. {-6+3y=1
{x+3y=-1
b. {-6+3y=3
{x+3y=-4
(i) Find the inverse of the (common) coefficient matrix of the two systems. A⁻¹=[]
(ii)Find the solutions to the two systems by using the inverse, i.e. by evaluating A⁻¹B where B represents the right hand side (i.e.
Previous question
B=[1 -1]for system (a) and B=[3 -4] for system (b))
solution to system (a):x= ,y=
solution to system (b):x= ,y=
Answer:
(i)
[tex]A^{-1}=\left[\begin{array}{cc}-\dfrac{1}{7}&\dfrac{1}{7}\\\\\dfrac{1}{21}&\dfrac{2}{7}\end{array}\right][/tex]
(ii) (a) x = -2/7, y = -5/21; (b) x = -1, y = -1
Step-by-step explanation:
Given the following systems of equations, you want the inverse of the coefficient matrix, and the solution to each system found by multiplying that coefficient matrix by the constant vector.
-6x +3y = 1x +3y = -1-6x +3y = 3x +3y = -4Inverse matrixThe calculator display in the attachment shows the coefficient matrix and its inverse. The inverse of a matrix is the transpose of the cofactor matrix, divided by the determinant. For a 2×2 matrix, the transpose of the cofactor matrix is simply the matrix obtained by swapping the diagonal elements, and negating the off-diagonal elements.
Here the determinant is (-6)(3) -(1)(3) = -21. So, the upper left element of the inverse matrix, for example, is 3/(-21) = -1/7, as shown in the attachment.
[tex]A^{-1}=\left[\begin{array}{cc}-\dfrac{1}{7}&\dfrac{1}{7}\\\\\dfrac{1}{21}&\dfrac{2}{7}\end{array}\right][/tex]
SolutionsMultiplying the inverse matrix (A⁻¹) by each constant column vector (B) gives a result that is a column vector. We can append the constant vectors to form a matrix of the two column vectors, saving a little work in computing the solutions to the two systems. The columns of the result are the solutions to the two systems.
system (a): x = -2/7, y = -5/21
system (b): x = -1, y = -1
__
Additional comment
The second attachment shows the use of an augmented matrix to find both the inverse of the coefficient matrix and the solutions to the systems of equations. The input is the coefficient matrix augmented by a 2×2 identity matrix and the two constant vectors. The output is the identity matrix, the the inverse of the coefficient matrix, and the two solution vectors.
<95141404393>
Use properties of logarithms to expand into a difference of logarithms. log 8 22/3
The logarithmic expression log₈(22/3) can be expanded into a difference of logarithms using properties of logarithms.
To expand the logarithmic expression log₈(22/3) into a difference of logarithms, we can apply the quotient rule of logarithms. According to the quotient rule, log base a of (b/c) is equal to log base a of b minus log base a of c. Applying this rule to the given expression, we get
log₈(22) - log₈(3).
This represents a difference of logarithms, where the numerator of the original expression becomes the first term and the denominator becomes the second term. Therefore, log₈(22/3) can be expanded as
log₈(22/3) = log₈(22) - log₈(3).
By applying properties of logarithms, we can simplify and manipulate logarithmic expressions, allowing us to break down complex expressions into simpler forms, which aids in calculations and problem-solving involving logarithms.
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the sides of a triangle are 10, 17 and 21 inches long. find
a) the smallest angle of the triangle
b) the diameter of the circumscribed circle
The smallest angle of the triangle is 25.46° and the diameter of the circumscribed circle is 23.31 inches.
Now the given sides are,
10, 17 and 21
Therefore, the angles we get,
tan θ = (10/17)
⇒θ = 25.46°
tan θ = (17/21)
⇒θ = 38.99°
tan θ = (17/10)
⇒θ = 59.53°
Hence, the smallest angle is 25.46°
Now for the diameter of the circumscribed circle,
if a, b, c are the lengths of the three sides of a triangle and A, B, C are the corresponding measures of the opposite angles respectively, then the ratio
a/sinA = b/sinB = c/sinC = d
is said to the length of the diameter of the circumscribed circle of the triangle.
So let a = 10 and A = 25.46°
⇒ d = 10/sin25.46°
⇒ d = 10/0.429
⇒ d = 23.31 inches
Hence, the smallest angle of the triangle is 25.46° and the diameter of the circumscribed circle is 23.31 inches.
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