How many pounds are in a kilogram

Therefore, the probability that either component A or B, but not both, fails is 0.5.

To find the probability that either component A or B, but not both, fails, we can use the concept of exclusive OR (XOR).

The XOR operation evaluates to true (1) if and only if exactly one of the conditions is true. In this case, we want to calculate the probability of either A or B failing, but not both.

Let's denote the event "A fails" as A and the event "B fails" as B. The probability that both A and B fail simultaneously is given as 0.1.

Now, let's break down the possible scenarios:

A fails, B does not fail: The probability of this event is P(A) * (1 - P(B)) = 0.4 * (1 - 0.5) = 0.4 * 0.5 = 0.2.

A does not fail, B fails: The probability of this event is (1 - P(A)) * P(B) = (1 - 0.4) * 0.5 = 0.6 * 0.5 = 0.3.

Since we are interested in the XOR scenario, where either A or B fails, but not both, we need to sum up the probabilities from scenarios 1 and 2:

P(A XOR B) = P(A fails, B does not fail) + P(A does not fail, B fails)

= 0.2 + 0.3

= 0.5.

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Step-by-step explanation:

the underlined numbers are the answers

This means that we can estimate, with 95% confidence, that the average time spent at all Taco Bell drive-throughs is between 166.1 seconds (173 - 6.9) and 179.9 seconds (173 + 6.9).

To estimate the average time spent at all Taco Bell drive-throughs with 95% confidence, we can use a confidence interval. The formula for the confidence interval is:

CI = x ± Z * (σ / √n)

Where:

x is the sample mean (173 seconds)

Z is the Z-score corresponding to the desired confidence level (95% confidence corresponds to a Z-score of approximately 1.96)

σ is the population standard deviation (35 seconds)

n is the sample size (100 cars)

Plugging in the values, we get:

CI = 173 ± 1.96 * (35 / √100)

Calculating the expression inside the parentheses, we have:

CI = 173 ± 1.96 * (35 / 10)

Simplifying further, we get:

CI = 173 ± 1.96 * 3.5

CI = 173 ± 6.86

Rounding to one decimal place, the confidence interval is:

CI = 173 ± 6.9 seconds

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H = {[x y] 5x²+6y² ≤ 1} represents the set of points on and inside an ellipse in the xy-plane. To show that H is not a subspace of R², we can provide two examples.

Example 1: Let's consider the vector [1 0]. Since 5(1)² + 6(0)² = 5, this vector satisfies the inequality 5x² + 6y² ≤ 1. However, if we multiply this vector by a scalar, say 2, we get [2 0]. Now, 5(2)² + 6(0)² = 20, which does not satisfy the inequality. Hence, the vector [2 0] is not in H, showing that H is not closed under scalar multiplication.

Example 2: Let's consider two vectors, [1 0] and [0 1]. Both vectors satisfy the inequality 5x² + 6y² ≤ 1. However, if we add these vectors together, [1 0] + [0 1] = [1 1]. Now, 5(1)² + 6(1)² = 11, which does not satisfy the inequality. Therefore, the vector [1 1] is not in H, demonstrating that H is not closed under vector addition.

In both examples, we have shown that H fails to satisfy the closure properties of a subspace. H is not closed under scalar multiplication in the first example, and it is not closed under vector addition in the second example. Hence, we can conclude that H is not a subspace of R².

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The price index for a train ticket from Lanchester to London, using the base year 2000, can be calculated by dividing the price in each year by the price in the base year and multiplying by 100. The price index for 1995 is 50, for 1996 is 66.67, for 1997 is 83.33, and for 2000 (the base year) is 100.

To calculate the price index for a train ticket from Lanchester to London, we need to compare the prices in different years to a base year, which in this case is 2000. The formula for calculating the price index is (Price in Year / Price in Base Year) x 100.

For the year 1995, the price of the ticket is £30. Dividing this price by the price in the base year (2000), which is £60, gives us 0.5. Multiplying this by 100 gives a price index of 50.

For 1996, the ticket price is £40. Dividing by the base year price (£60) gives us 0.6667. Multiplying by 100 gives a price index of approximately 66.67.

Similarly, for 1997, the price index is 83.33, as the ticket price of £50 divided by the base year price (£60) gives us 0.8333 when multiplied by 100.

Lastly, for the base year 2000, the price index is 100, as it is the reference point and used as the denominator in the formula.

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The equation of the plane containing the lines L₁ and L₂ is -2x - y + z + 3 = 0.

To find the equation for the plane containing the lines L₁ and L₂, we can use the cross product of the direction vectors of the two lines.

The direction vector of L₁ is (1, 1, 1), and the direction vector of L₂ is (1, -1, 0). Taking the cross product of these two vectors will give us a vector that is orthogonal (perpendicular) to both lines and therefore normal to the plane.

Let's calculate the cross product:

N = (1, 1, 1) × (1, -1, 0)

To calculate the cross product, we can use the determinant method:

N = (1 * (-1) - 1 * 1, 1 * 0 - 1 * 1, 1 * 1 - 1 * 0)

= (-2, -1, 1)

Now, we have the normal vector N = (-2, -1, 1) which is orthogonal to the plane containing L₁ and L₂.

Next, we need to find a point on the plane. We can choose any point on either of the lines L₁ or L₂. Let's choose a point on L₁. When t = 0, the parametric equation for L₁ gives us the point (1, 0, -1).

Now, we have a point (1, 0, -1) on the plane and the normal vector N = (-2, -1, 1) orthogonal to the plane. We can use the point-normal form of the equation for a plane to find the equation of the plane.

The point-normal form of the equation of a plane is:

N · (P - P₀) = 0

where N is the normal vector, P is a point on the plane, and P₀ is a known point on the plane.

Substituting the values we have:

(-2, -1, 1) · ((x, y, z) - (1, 0, -1)) = 0

Simplifying:

-2(x - 1) - (y - 0) + (z + 1) = 0

-2x + 2 - y + z + 1 = 0

-2x - y + z + 3 = 0

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When T = 170° and P = 20 lb/in², the volume of the gas is approximately 119.0 in³ (option b).

The problem states that the volume V of a gas varies directly with the temperature T and inversely with the pressure P. Mathematically, this can be represented as V = k(T/P), where k is the constant of variation.

To solve the problem, we can use the given values of V, T, and P to find the value of k. We have V = 350.0 in³, T = 250°, and P = 10 lb/in². Plugging these values into the equation V = k(T/P), we can solve for k.

350.0 = k(250/10)

35 = k(25)

k = 35/25 = 1.4

Now that we have the value of k, we can find the volume V when T = 170° and P = 20 lb/in².

V = k(T/P) = 1.4(170/20) = 11.9 in³

Rounding to the nearest tenth, the volume of the gas is approximately 119.0 in³, which corresponds to option b.

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Answer:

(a) Median

Step-by-step explanation:

The median is least affected by the addition of an outlier because it concerns the values in the middle of a sorted data list. Adding an extremely high number would not affect the middle data, so the median would not really be influenced by the outlier. On the other hand, the mean and standard deviation take into account all the values in the data, so adding an extreme high outlier would change the mean and standard deviation. The range and maximum are also affected since they involve the highest and lowest values, and the highest value would be different. Therefore, the median is the least affected if an extreme high outlier is added to the data.

The time spent unsupervised by the four year-olds in the country is normally distributed with a standard deviation of 1.8 hours. If we randomly select a four year-old who resides in a rural area, we would like to know the time he or she spends alone each day. Part (a) : Here we are supposed to find the mean of time spent alone by a child from a rural area.

Now to find `μ` we substitute the values in the formula and simplify it as follows: `-

1.28 = (3 - μ) / 1.8`.

Now we solve for

`μ` i.e,

Mean of time spent alone by a child from a rural area:

`μ = 3 + 1.8 (1.28)` `μ = 5.184`

Thus, the mean time that a four year-old from a rural area spends alone is `5.184` hours per day.

Part (b) : Here we are supposed to find the standard deviation of time spent alone by a child from a rural area.

Now to find `σ` we substitute the values in the formula and simplify it as follows:

`-1.28 = (x - 5.184) / σ`

Now we solve for `σ` i.e, standard deviation of time spent alone by a child from a rural area:

`σ = (x - 5.184) / -1.28`

Therefore, the standard deviation of time spent alone by a four year-old from a rural area is `(x - 5.184) / -1.28`Part (c) : Here we need to find the probability that a 4-year-old from a rural area spends more than 2 hours alone.

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Mean: The formula for calculating mean is [tex]\[\overline{x}=\frac{\sum\limits_{i=1}^{n}x_{i}}{n}\][/tex]

Variance: The formula for calculating variance is

Given distribution is:1 I 3 4 7 8

P(X=r)0.30.20.1

The mean and variance can be calculated as follows:

Mean calculation:Now, calculating the mean of distribution; we have:\

The variance of the distribution is 1.58.

Summary: The mean of the distribution is 1.3 and the variance of the distribution is 1.58.

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If the investment grows by 50% every 20 years, we can use the formula for exponential growth to express the balance, B, t years later.

The formula for exponential growth is given by: B = P * (1 + r)^t.Where: B is the balance after t years.P is the initial investment (principal). r is the growth rate per year.In this case, the initial investment is $100 and the growth rate is 50%, which can be written as 0.5. The time period is t years. Substituting the values into the formula, we get:B = 100 * (1 + 0.5)^t . Simplifying further: B = 100 * (1.5)^t.

Therefore, the formula that expresses the balance, B, t years later is:

B = 100 * (1.5)^t.

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A confidence interval can be used to define a range of plausible values for an unknown parameter, like the variance ratio.

variances of two portfolios with sample variances of s1^2 and s2^2. Let's calculate the confidence interval for the ratio of population variances 05 using the given information.

[tex](s1^2 / s2^2) * (Fα/2),v2, v1 ≤ (s1^2 / s2^2) * (F1-α/2),v1,v2[/tex]

freedom. We must first determine the degrees of freedom and find the critical values for the F-distribution.In this case, v1 = n1 - 1 = 15 - 1 = 14 and v2 = n2 - 1 = 15 - 1 = 14.Using a significance level of 0.05 and degrees of freedom (14,14), the critical values from the F-distribution table are 0.414 and 2.377. Thus, the 95% confidence interval for the variance ratio is calculated as follows:

[tex](s1^2 / s2^2) * (Fα/2),v2, v1 ≤ (s1^2 / s2^2) * (F1-α/2),v1,v2= (0.0049 / 0.0064) * (2.377) ≤ (0.0049 / 0.0064) * (0.414)= 1.8375 ≤ 1.2156[/tex]

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The limit does not exist when X approaches infinity. So, the answer is X = 0.

Given lim (x³e³) / X, with X approaching infinity.

We need to apply L'Hopital's rule.

As this expression is of the form infinity / infinity

So, differentiate the numerator and denominator with respect to X.

d/dx (x³e³) / d/dx (X) = 3x² e³ / 1d/dx (X)

= 1

Now we get the expression lim (3x² e³) / 1 with X approaching infinity

This still gives infinity / infinity, hence we again apply L'Hopital's rule.

d/dx (3x² e³) / d/dx (X)

= 6xe³ / 1

Now we get the expression lim (6xe³) / 1 with X approaching infinity

This expression evaluates to infinity as the numerator is infinity while the denominator is a finite number.

Thus, the limit does not exist when X approaches infinity.

So, the answer is X = 0.

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To calculate the pooled variance for the hypothesis test, assuming equal variances, we need to combine the variances from the two groups.

The pooled variance is used when the assumption of equal variances between the two groups is reasonable. The formula for pooled variance is: Pooled Variance = ((n1 - 1) * S1^2 + (n2 - 1) * S2^2) / (n1 + n2 - 2) where n1 and n2 are the sample sizes of the two groups, and S1^2 and S2^2 are the variances of the respective groups.

Using the given information:

Group 1 (Glucose): n1 = 15, S1 = 20.3

Group 2 (Placebo): n2 = 15, S2 = 13.5

Calculating the pooled variance:

Pooled Variance = ((15 - 1) * 20.3^2 + (15 - 1) * 13.5^2) / (15 + 15 - 2)

After performing the calculations, the pooled variance is found to be approximately 208.293 when rounded to three decimal places.

The pooled variance for the hypothesis test, assuming equal variances, is approximately 208.293. This value represents the combined variance of the two groups and is used in the calculation of the test statistic in the two-sample t-test.

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The set spanned by the function f(x) = a + bx + cx² + dx³, given f(2) = 0 and f'(-2) = 0, consists of all polynomials of degree 3 or less that satisfy these conditions.

The conditions f(2) = 0 and f'(-2) = 0 imply that the polynomial passes through the point (2, 0) and has a horizontal tangent at x = -2. To find the set spanned by this function, we need to determine the coefficients a, b, c, and d that satisfy these conditions.

By solving the equations, we can obtain a unique polynomial that meets these criteria. Therefore, the set spanned by the given function is a single polynomial of degree 3 or less that satisfies the conditions f(2) = 0 and f'(-2) = 0.


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The weighted mean medical payments for these five plants is $11536.90

From the question, we have the following parameters that can be used in our computation:

The table of values

The weighted mean medical payments for these five plants is calculated as

Weighted mean = Sum/Count

So, we have

Weighted mean = (7,773 * 125 + 14,808 * 404 + 12,392 * 253 + 6,708 * 105 + 3,532 * 70)/(125 + 404 + 253 + 105 + 70)

Evaluate

Weighted mean = 11536.90

Hence, the weighted mean is $11536.90

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Question

A corporation that operates five suppliers of athletic apparel in a region provides merchandise for a shoe company. The shoe company recently sought information from the five plants. One variable for which data were collected was the total money​ (in dollars) the company spent on medical support for its employees in the first three months of the year. Data on number of employees at the plants are also shown below.

Data:

Medical Employees

$7,773 125

$14,808 404

$12,392 253

$6,708 105

$3,532 70

Compute the weighted mean medical payments for these five plants using the numbers of employees as the weights.

The answer in the form a + b i  is :

18⁽⁴π/³⁾ * (6⁽²π/³⁾* ²) * (cos(7π/9) + i sin(7π/9))

To simplify the expression using De Moivre's theorem, we need to evaluate the expression 18√6(cos(4π/3) + i sin(4π/3)).

De Moivre's theorem states that for any complex number

z = r(cosθ + i sinθ), its nth power can be expressed as

zⁿ = rⁿ (cos(nθ) + i sin(nθ)).

In this case, we have z = 18√6 and

n = 4π/3.

Let's calculate the simplified form:

r = 18√6

θ = 4π/3

Using De Moivre's theorem, we can rewrite the expression as:

18√6(cos(4π/3) + i sin(4π/3)) = (18√6)⁽⁴π/³⁾ (cos(4π/3 * 4π/3) + i sin(4π/3 * 4π/3))

Now, let's simplify the expression further:

(18√6)⁽⁴π/³⁾ = 18⁽⁴π/³⁾  * (6^⁽¹/²⁾)⁽⁴π/³⁾ = 18⁽⁴π/³⁾ * (6⁽⁴π/⁶⁾) = 18⁽⁴π/³⁾ * (6⁽⁽²π/³⁾ * ²⁾)

cos(4π/3 * 4π/3) = cos(16π/9) = cos(2π + 7π/9)

= cos(7π/9)

sin(4π/3 * 4π/3) = sin(16π/9) = sin(2π + 7π/9)

= sin(7π/9)

Putting it all together, the simplified expression is:

18⁽⁴π/³⁾* (6⁽⁽²π/³⁾ * ²⁾) * (cos(7π/9) + i sin(7π/9))

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Step-by-step explanation:

My z-score tables are set up to show the area to the LEFT
  so you will need to find the z-score that is   1-.10 = .90  

  which , by looking at the tables is z-score =  +1.28

The length of the curve y = x^(3/2), 0≤x≤2 is 4√5 units.

The length of the curve y = ln(sec(x)), 0≤x≤π/4 is ln(√2+1) units.

To find the length of a curve, we can use the formula for arc length:

L = ∫(a to b) √(1 + (dy/dx)^2) dx.

For the curve y = x^(3/2), we first find dy/dx = (3/2)x^(1/2). Plugging this into the arc length formula, we have:

L = ∫(0 to 2) √(1 + (3/2)^2x) dx = ∫(0 to 2) √(1 + 9/4 x) dx.

Simplifying the expression inside the square root and integrating, we get:

L = (4/5) (1 + (9/4)^(3/2)) = 4√5.

For the curve y = ln(sec(x)), we find dy/dx = tan(x). Plugging this into the arc length formula, we have:

L = ∫(0 to π/4) √(1 + tan^2(x)) dx = ∫(0 to π/4) √sec^2(x) dx.

Simplifying the expression inside the square root and integrating, we get:

L = ∫(0 to π/4) sec(x) dx = ln(√2 + 1).

Therefore, the length of the curve y = x^(3/2), 0≤x≤2 is 4√5 units, and the length of the curve y = ln(sec(x)), 0≤x≤π/4 is ln(√2 + 1) units.

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The initial population at time t = 0 was 40,000.

To determine the initial population at time t = 0, we can use the concept of doubling time and the given information.

The doubling period refers to the time it takes for a population to double in size. In this case, the doubling period is stated as 20 minutes.

Let's denote the initial population as P0. We know that after 20 minutes (one doubling period), the population becomes twice its initial size. So, we can set up the following equation:

P0 * 2 = 80000

Now, we can solve for P0:

P0 = 80000 / 2

P0 = 40000

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The area of surface of revolution is 401.354 square units (correct to 3 decimal places).

a) Arc Length:

First, we need to find the derivative of the given function y= 6x + 3 to get dy/dx, then substitute into the following equation:

Let L be the length of the curve, so,  So the length of the curve is 18.438 units (correct to 3 decimal places).

b) Area of surface of revolution:

The formula to find the surface area of revolution:

Substituting values in the above formula, we get:

Solving this integral with x = 6 gives an area of surface of revolution is 401.354 square units (correct to 3 decimal places).

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To construct a confidence interval for the proportion of all customers who experienced an interruption, we can use the formula:

Confidence Interval = Point Estimate ± (Critical Value) * (Standard Error)

First, let's calculate the point estimate:

Point Estimate = Number of customers who experienced an interruption / Total sample size

Point Estimate = 75 / 550 ≈ 0.136 (rounded to three decimal places)

Next, we need to determine the critical value corresponding to a 98% confidence level. Since the sample size is large (n = 550) and the data are assumed to be approximately normally distributed, we can use the z-table.

The critical value for a 98% confidence level can be found by finding the z-score that corresponds to an area of (1 - 0.98) / 2 = 0.01 in the upper tail of the standard normal distribution. Using the z-table, this critical value is approximately 2.33 (rounded to two decimal places).

Next, we need to calculate the standard error:

Standard Error = sqrt((Point Estimate * (1 - Point Estimate)) / Sample Size)

Standard Error = sqrt((0.136 * (1 - 0.136)) / 550) ≈ 0.016 (rounded to three decimal places)

Now we can construct the confidence interval:

Confidence Interval = 0.136 ± (2.33 * 0.016)

Confidence Interval ≈ 0.136 ± 0.037 (rounded to three decimal places)

So the 98% confidence interval for the proportion of all customers who experienced an interruption is approximately 0.099 < p < 0.173.

c. The company's quality control manager claims that no less than 10% of its customers experienced an interruption. We can check if this claim is contradicted by examining the confidence interval.

In the confidence interval, the lower bound is 0.099, which is greater than 0.1. Therefore, the confidence interval does not contradict the claim. It is possible that at least 10% of the customers experienced an interruption based on the provided data.

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17/36

Step-by-step explanation:

If t he question was check the probability of only red so it would be 12/36 but here it is red and orange so we will add the both 12 + 5 which will be 17 so 17 /36 is the right answer.

Answer:  [tex]\frac{17}{36}[/tex] or 0.472 or 47.2%

Step-by-step explanation:

P( red skittle OR orange skittle)=[tex]\frac{12}{36} +\frac{5}{36}[/tex]

P( red skittle OR orange skittle)=[tex]\frac{17}{36}[/tex] or 0.472 or 47.2%

The firm's weighted average cost of capital (WACC) is 8.165%. The closest option given is B. 5.44%, but none of the provided options match the calculated value exactly.

To calculate the weighted average cost of capital (WACC), we need to consider the cost of debt and the cost of equity.

1. Cost of Debt:

The cost of debt can be determined using the bond yield. Given that the bonds are selling at 98% of face value, the market price of each bond is $980. The annual coupon payment is 7% of $1,000, which is $70. The bonds mature in 7.5 years, so we can calculate the yield to maturity (YTM) using financial calculators or spreadsheet software. Let's assume the YTM is 6%.

2. Cost of Equity:

The cost of equity can be calculated using the dividend discount model (DDM). The dividend growth rate is 2%, and the next year's dividend is expected to be $1.55 per share. The market price of each share is $15. Using the DDM formula:

Cost of Equity = Dividend / Stock Price + Growth Rate

Cost of Equity = $1.55 / $15 + 2% = 0.1033 or 10.33%

3. Weighted Average Cost of Capital (WACC):

To calculate the WACC, we need to consider the weight of debt and equity in the capital structure. Since the question does not provide information about the proportion of debt and equity, we'll assume an equal weighting of 50% for both.

WACC = (Weight of Debt * Cost of Debt) + (Weight of Equity * Cost of Equity)

WACC = (0.5 * 6%) + (0.5 * 10.33%)

WACC = 0.03 + 0.05165 = 0.08165 or 8.165%

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Let us say two different inputs x₁ and x₂ so that f(x₁) = f(x₂).

f(x₁) = f(x₂) implies:

(e^x₁ - e^(-x₁)) / (e^x₁ + e^(-x₁)) = (e^x₂ - e^(-x₂)) / (e^x₂ + e^(-x₂))

(e^x₁ - e^(-x₁))(e^x₂ + e^(-x₂)) = (e^x₂ - e^(-x₂))(e^x₁ + e^(-x₁))

(e^x₁e^x₂ + e^x₁e^(-x₂) - e^(-x₁)e^x₂ - e^(-x₁)e^(-x₂)) = (e^x₂e^x₁ + e^x₂e^(-x₁) - e^(-x₂)e^x₁ - e^(-x₂)e^(-x₁))

e^x₁e^(-x₂) - e^(-x₁)e^x₂ = e^x₂e^(-x₁) - e^(-x₂)e^x₁

ln(e^x₁e^(-x₂) - e^(-x₁)e^x₂) = ln(e^x₂e^(-x₁) - e^(-x₂)e^x₁)

Using the properties of logarithms:

x₁ - x₂ = x₂ - x₁

This simplifies to:

0 = 0

The equation zero = zero is proper for any x₁ and x₂. This implies that the idea that f(x₁) = f(x₂) leads to a contradiction. Hence, the characteristic f(x) = (e^x - e^(-x)) / (e^x + e^(-x)) is injective or one-to-one.

Thus, as the feature is one-to-one, it means that it has an inverse.

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Your question seems incomplete, the probable complete question is:

Examine whether the function f(x)=e^x-e^-x/e^x+e^-x is inverse or not.

The correct answer is not provided in the options. None of the given options match the calculated probability.

P(Headache)

To calculate the probability of having a headache, we need to divide the number of workers who reported a headache by the total number of workers. According to the table, 33 workers reported a headache.

P(Headache) = 33 / 167 ≈ 0.197

Therefore, the correct answer is not provided in the options. None of the given options match the calculated probability.

ii) P(Shortness of Breath or Night)

To calculate the probability of having either shortness of breath or working the night shift, we need to add the number of workers who reported shortness of breath and the number of workers who worked the night shift, and then divide by the total number of workers.

According to the table, 25 workers reported shortness of breath, and 38 workers worked the night shift.

P(Shortness of Breath or Night) = (25 + 38) / 167 ≈ 0.323

Therefore, the correct answer is option B) 0.323.

iii) P(Evening | Weakness)

To calculate the probability of working the evening shift given that the worker has weakness symptoms, we need to divide the number of workers who worked the evening shift and had weakness symptoms by the total number of workers with weakness symptoms.

According to the table, 16 workers had weakness symptoms, and out of those, 16 worked the evening shift.

P(Evening | Weakness) = 16 / 16 = 1

Therefore, the correct answer is not provided in the options. None of the given options match the calculated probability.

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Answer:

it will take Tyrell and Hunter approximately 2.4 hours to mow the lawn together.

Step-by-step explanation:

To determine the number of hours it will take for Tyrell and Hunter to mow the lawn together, we can use the concept of work rates.

Tyrell mow 1/4 per hour

Hunter mow 1/6 per hour

When working together, their work rates are additive. Therefore, the combined work rate of Tyrell and Hunter is:

1/4 + 1/6 = 3/12 + 2/12 = 5/12 of the lawn per hour.

(5/12) * x = 1 (since they need to mow the entire lawn, which is equivalent to 1).

x = (1) * (12/5) = 12/5 = 2.4 hours.

The expression arccos(cos(πº)) can be evaluated. The result is π radians. The expression arccos(cos(270º)) can be90º or π/2 radians. The expression cos(arccos(2)) cannot be evaluated.

The expression arccos(cos(πº)) can be evaluated because it involves the composition of inverse cosine and cosine functions. The cosine of πº is -1, and the arccosine function returns the angle whose cosine is -1. Since the cosine function has a period of 2π, the angle π is equivalent to 180º, and therefore arccos(cos(πº)) = π.

Similarly, the expression arccos(cos(270º)) can be evaluated. The cosine of 270º is 0, and the arccosine function returns the angle whose cosine is 0. In this case, the angle is 90º or π/2 radians.

The expression cos(arccos(2)) cannot be evaluated because the arccosine function only returns values between -1 and 1. Since 2 is out.

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1. The probability that the high temperature will exceed 100 is  0.0228 or 2.28%.

2. To satisfy quality control requirements, the standard deviation (σ) of the battery production needs to be determined such that the probability of the voltage being between 1.45V and 1.52V is at least 99%.

1. To find the probability that the high temperature will exceed 100, we can standardize the variable using the z-score formula: z = (x - u) / a, where x is the value of interest, u is the mean, and a is the standard deviation. Then, we can calculate the probability using the standard normal distribution table or software. For this case, we need to find P(X > 100), where X is the high temperature. By standardizing, we get z = (100 - 87.2) / 6.4 ≈ 2.0. Consulting the standard normal distribution table, we find that the probability is approximately 0.0228 or 2.28%.

2. To determine the required standard deviation for the battery production, we need to find the z-scores corresponding to the lower and upper voltage limits (1.45V and 1.52V) using the formula z = (x - V) / σ, where V is the mean voltage and σ is the standard deviation. We want the probability of the voltage falling within this range to be at least 99%, so we need to find the z-scores that give the cumulative probability of 0.99 and (1 - 0.99)/2 = 0.005 on each tail. Once we have the z-scores, we can rearrange the formula to solve for σ. This calculation requires the use of the standard normal distribution table or software.

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Given (X,Y) has the two dimensional Gaussian distribution with parameters: vector of expectations μ = (EX,EY) = (1,-1) and covariance matrix as shown below: Cov(X,X) Cov(X,Y) 4 2 C = Cov(Y,X) Cov(Y,Y) 2 2.

We are supposed to find the correlation coefficient of X and Y.

Recall that the correlation coefficient is given by the formula

Cor(X, Y) = Cov(X, Y) / (SD(X) * SD(Y)) ... (1)

Thus, we will need to calculate Cov(X, Y), SD(X) and SD(Y).

Summary: The correlation coefficient of X and Y for the given two-dimensional Gaussian distribution is 1/√2.

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