A business has a savings account that earns a 3% annual interest rate. At the end of 1996, the business had $4,000 in the account. The


formula F


+


100


is used to determine the amount in the savings account.


p(1


a)
. Fis the final amount,


⢠p is the initial investment amount,
. Ris the annual interest rate, and
. Tis the time in years.


To the nearest dollar, how much did the business initially invest in 1991?

The solution set of the inequality 3x-2 < 2-2x is:

(4/5, ∞)

To find this we need to isolate the variable in the inequality.

Here we have:

3x - 2 < 2 - 2x

add 2x in both sides and add 2 in both sides, then we will get:

3x + 2x < 2 + 2

5x < 4

Now we can divide both sides by 5 to get:

x < 4/5

That is the inequality solved.

Then the solution set of the inequality is:

(4/5, ∞)

The set of all real numbers larger than 4/5.

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To find the total volume of the silo, we need to add the volume of the hemisphere on top to the volume of the cylinder at the bottom.

The volume of a hemisphere is given by:

V_hemi = (2/3)πr^3

where r is the radius of the hemisphere.

Substituting r = 9ft, we get:

V_hemi = (2/3)π(9ft)^3

= 1521π ft^3

The volume of a cylinder is given by:

V_cyl = πr^2h

where r is the radius of the cylinder and h is its height.

Substituting r = 9ft and h = 35ft, we get:

V_cyl = π(9ft)^2(35ft)

= 2673π ft^3

Therefore, the total volume of the silo is:

V_silo = V_hemi + V_cyl

= 1521π + 2673π

= 4194π ft^3

≈ 13160 ft^3

Rounding to the nearest cubic foot, the silo can hold approximately 13160 cubic feet of grain.

The critical points and their classifications are:

(0, 1) - saddle point

(0, -1) - saddle point

(6, 1) - local minimum

(6, -1) - local minimum

To find the critical points of the function f(x, y) = x³ + y³ – 9x² – 3y - 6, we need to find the points where the partial derivatives of f with respect to x and y are zero.

∂f/∂x = 3x² - 18x = 3x(x - 6)

∂f/∂y = 3y² - 3 = 3(y² - 1)

Setting these partial derivatives equal to zero and solving for x and y, we get:

x = 0 or x = 6

y = ±1

So the critical points are (0, 1), (0, -1), (6, 1), and (6, -1).

To classify each critical point, we need to compute the second partial derivatives of f:

∂²f/∂x² = 6x - 18

∂²f/∂y² = 6y

∂²f/∂x∂y = 0

At (0, 1):

∂²f/∂x² = -18 < 0 (concave down)

∂²f/∂y² = 6 > 0 (concave up)

So (0, 1) is a saddle point.

At (0, -1):

∂²f/∂x² = -18 < 0 (concave down)

∂²f/∂y² = 6 > 0 (concave up)

So (0, -1) is a saddle point.

At (6, 1):

∂²f/∂x² = 18 > 0 (concave up)

∂²f/∂y² = 6 > 0 (concave up)

So (6, 1) is a local minimum.

At (6, -1):

∂²f/∂x² = 18 > 0 (concave up)

∂²f/∂y² = 6 > 0 (concave up)

So (6, -1) is a local minimum.

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Answer:

? equals 80

Step-by-step explanation:

When the points are rotated 90 degrees clockwise about the origin, the result is:

To rotate a point 90 degrees clockwise about the origin, you can use the following rule: (x, y) becomes (y, -x). Let's apply this rule to the given points:

I - (3, 1)

Rotated I: (1, -3)

J - (5, -1)

Rotated J: (-1, -5)

H - (3, -3)

Rotated H: (-3, -3)

So, after a 90-degree clockwise rotation about the origin, the new coordinates of the points are:

I: (1, -3)

J: (-1, -5)

H: (-3, -3)

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The functions that represents the cost are

(a) y = 8800, (b) y = 1900 + 4/7x and (c) y = 4800, x ≤ 150;  y = 1200 + 24x x > 150

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

The graph

The function (a) is a horizontal line that passes through y = 8800

So, the function is

y = 8800

The function (b) is a linear function that passes through

(0, 1900) and (175, 2000)

So, the function is

y = 1900 + 4/7x

The function c is a piecewise function with the following properties

So, the function is

y = 4800, x ≤ 150

y = 1200 + 24x x > 150

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The relative frequency for selecting a teen out of all the people who like chocolate is calculated by dividing the number of teens who like chocolate (N) by the total number of people who like chocolate (T).

To find the relative frequency for selecting a teen out of all the people who like chocolate, you need to follow these steps:

Step 1: Determine the total number of people who like chocolate (let's call this T).
Step 2: Determine the number of teens who like chocolate (let's call this N).
Step 3: Calculate the relative frequency by dividing the number of teens who like chocolate (N) by the total number of people who like chocolate (T).

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In triangle ABC with a right angle at B, to make angle A greater than 75 degrees, you can choose BC = 1 unit and hypotenuse AC = 3 units.

In a right-angled triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. In our case, sin(A) = BC/AC. To make angle A greater than 75 degrees, we need sin(A) > sin(75). Using a calculator, sin(75) ≈ 0.9659. So, we need BC/AC > 0.9659.

Let's take BC = 1 unit, then we need AC > 1/0.9659 ≈ 1.035 units. To keep it simple, we can choose AC = 3 units. Now, sin(A) = 1/3 ≈ 0.3333, and the corresponding angle A is around 19.47 degrees. Note that this is greater than 75 degrees, fulfilling the requirement.

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After 2 hours, approximately 4.38 grams of Francium remain from the 480-gram sample.

What is Algebraic expression ?

An algebraic expression is a combination of variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. It may contain one or more terms, with each term separated by a plus or minus sign. Algebraic expressions are used in algebra to represent mathematical relationships and formulas.

a) To write an exponential function that models the mass of Francium remaining after t minutes, we can use the formula:

N = N0 * [tex](1/2)^{(t / t1/2)}[/tex]

where N is the amount remaining after time t, N0 is the initial amount, t1/2 is the half-life, and (t/t1/2) means raised to the power of t/t1/2.

In this case, the initial amount is 480 grams, the half-life is 22 minutes, and we want to find the amount remaining after t minutes. Therefore, the exponential function that models the mass of Francium remaining after t minutes is:

N = 480 * [tex](1/2)^{t/22}[/tex]

b) 2 hours is equal to 120 minutes. To find how many grams of Francium remain after 2 hours, we can substitute t = 120 into the exponential function we found in part a):

N = 480 *[tex](1/2)^{ (120 / 22) }[/tex] ≈ 4.38 grams

Therefore, after 2 hours, approximately 4.38 grams of Francium remain from the 480-gram sample.

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The sample space for the number solid is {1, 2, 3, 4, 5, 6} and the probability of rolling 1 is 1/6.

The sample space refers to the set of all possible outcomes of an experiment, while a sample value is a specific outcome in the sample space.

For the given 8-sided solid, the sample space would be {1, 2, 3, 4, 5, 6}, as the faces labeled "skip" are not counted as sample values.

Now, let's calculate the probability of rolling a 1. Probability is the likelihood of a particular outcome occurring, which can be calculated by dividing the number of successful outcomes (in this case, rolling a 1) by the total number of possible outcomes.

The total number of possible outcomes is 6 (1, 2, 3, 4, 5, and 6). There is only one successful outcome: rolling a 1.

So, the probability of rolling a 1 is:

P(1) = (Number of successful outcomes) / (Total number of possible outcomes)
P(1) = 1 / 6

Thus, the probability of rolling a 1 on this 8-sided solid is 1/6 or approximately 0.1667, or 16.67%.\

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The wavelength of the light is approximately 0.687 μm.

Using the single slit diffraction formula, we have:

sin θ = (mλ) / w

where m is the order of the minimum, λ is the wavelength of the light, and w is the width of the slit.

We can rearrange the formula to solve for the wavelength of the light:

λ = (w sin θ) / m

Plugging in the given values, we get:

λ = (1.2 μm)(sin 32.3°) / 1 = 0.687 μm

Therefore, the wavelength of the light is approximately 0.687 μm.

The wavelength is the distance between two consecutive peaks or troughs in a wave. It is typically represented by the Greek letter lambda (λ) and is measured in meters or other units of length. The wavelength is an important characteristic of any wave, as it determines many of its properties, such as its speed and frequency.

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Therefore, the maximum value of f(x,y) inside the triangle is 80/9, which occurs along the line y = (-1/2)x + 4 at the point (8/3, 10/3), and the minimum value is -32, which occurs at the critical point (-8,4).

To find the maximum and minimum values of f(x,y) = xy on the region inside the triangle whose vertices are (6,2), (0,3), and (6,0), we use the method of Lagrange multipliers.

First, we need to find the critical points of f(x,y) subject to the constraint that (x,y) lies inside the triangle. We can express this constraint using the equations of the lines that form the sides of the triangle:

y = (-1/2)x + 4

y = (3/2)x

y = 0

Next, we set up the Lagrange multiplier equation:

∇f = λ∇g

where g(x,y) is the equation of the constraint, i.e., the triangle.

We have:

f(x,y) = xy

∇f = <y, x>

g(x,y) = y - (-1/2)x - 4 = 0

∇g = <-1/2, 1>

Setting ∇f = λ∇g, we get:

y = (-1/2)λ

x = λ

Substituting these into the constraint equation, we get:

(-1/2)λ - 4 = 0

Solving for λ, we get:

λ = -8

Substituting this into y = (-1/2)λ and x = λ, we get:

x = -8 and y = 4

Therefore, the only critical point of f(x,y) inside the triangle is (-8,4).

Next, we need to check the values of f(x,y) at the vertices and along the sides of the triangle.

At the vertices:

f(6,2) = 12

f(0,3) = 0

f(6,0) = 0

Along the line y = (3/2)x:

f(x, (3/2)x) = (3/2)x^2

Using the vertex (6,2) and the x-intercept (4/3, 2), we can see that the maximum value of (3/2)x^2 on this line occurs at x = 4. Therefore, the maximum value of f(x,y) along this line is:

f(4,6) = 24

Along the line y = (-1/2)x + 4:

f(x, (-1/2)x + 4) = (-1/2)x^2 + 4x

Using the vertex (6,2) and the x-intercept (8,0), we can see that the maximum value of (-1/2)x^2 + 4x on this line occurs at x = 8/3. Therefore, the maximum value of f(x,y) along this line is:

f(8/3,10/3) = 80/9

Finally, we need to check the values of f(x,y) at the critical point (-8,4). We have:

f(-8,4) = -32

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The derivative of the vector function is: r'(t) = (-2t/(7-t^2)) i + (1/(2sqrt(13+t))) j - 36e^(9t) k

We are given a vector function r(t) = ln(7-t^2)i + sqrt(13+t)j – 4e^(9t)k, and we need to find its derivative r'(t).

The derivative of a vector function is obtained by differentiating each component of the vector function separately.

So, let's differentiate each component:

r(t) = ln(7-t^2)i + sqrt(13+t)j – 4e^(9t)k

r'(t) = (d/dt) ln(7-t^2) i + (d/dt) sqrt(13+t) j - (d/dt) 4e^(9t) k

Using the chain rule of differentiation, we have:

r'(t) = -2t/(7-t^2) i + 1/(2sqrt(13+t)) j - 36e^(9t) k

Therefore, the derivative of the vector function is:

r'(t) = (-2t/(7-t^2)) i + (1/(2sqrt(13+t))) j - 36e^(9t) k

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The correct option is: (d) a discontinuity at x = 5 only.

The function f(x) is defined as:

[tex]f(x) = (x-3)/(x^2 - 8x + 15)[/tex]

The denominator of this function can be factored as:

[tex]x^2 - 8x + 15 = (x - 3)(x - 5)[/tex]

So the function can be rewritten as:

f(x) = (x - 3)/[(x - 3)(x - 5)]

Simplifying this expression, we get:

f(x) = 1/(x - 5)

Now it is clear that the function has a discontinuity at x = 5, since the denominator of the simplified expression becomes zero at that point.

Therefore, the correct option is:

d. a discontinuity at x = 5 only

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Line x is perpendicular to line y. Line z crosses lines x and y. Only statements 3 and 4 are true.

∠6 = ∠8 is not true because they both lie on the same plane and makes an angle of 180° and can never be true. ∠6 = ∠1 is also not true because ∠1 is clearly obtuse angle and ∠6 is clearly acute angle so they cannot be equal. Hence, statement a and b are false.

∠7 = ∠3 is always true because they are corresponding angles and corresponding angles are always equal. m∠2 + m∠4 = 180° is also true because they lie on same plane and have common vertex and hence, they are supplementary angles and make a sum of 180°.  Hence, statement 3 and 4 is always true.

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The correct option is option 3.

Based on the given integral, region D is in the first quadrant, and its boundaries are not explicitly given. However, we can deduce the boundaries of D by looking at the integrand. Since the integrand is f(x,y), we can see that we are integrating over the entire region D, which means that D must be the rectangle that contains all the other regions mentioned in the options.

Therefore, option 1 is not correct, as D is not bounded from the right by x=2, but rather extends indefinitely to the right. Option 2 is also not correct, as D extends beyond the line y=x. Option 4 is not correct either, as D is not bounded above by y=x^2, but rather extends beyond it. Options 5 and 6 are also not correct, as D extends beyond the lines y=0 and y=2.

Therefore, the correct option is option 3, which states that D is the region in the first quadrant that is bounded from the left by the line x=0. This is correct, as D extends indefinitely to the right, and is bounded from the left by x=0.

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Answer:

we 1st can get the weight of rat by

1 rat and 1 cat + 1 dog and rat = 30

2 rat + 1 cat + 1 dog = 30

Then 1 rat and cat measure 24 so

2 rat + 24 =30

2 rat + 24 =30 1 rat = 3 kg

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 10

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 101 cat + 3kg = 10

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 101 cat + 3kg = 101 cat = 7kg and

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 101 cat + 3kg = 101 cat = 7kg and 1 dog + 1 rat = 20 kg

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 101 cat + 3kg = 101 cat = 7kg and 1 dog + 1 rat = 20 kg 1 dog + 3kg = 20 kg

2 rat + 24 =30 1 rat = 3 kg 1 cat + 1 rat = 101 cat + 3kg = 101 cat = 7kg and 1 dog + 1 rat = 20 kg 1 dog + 3kg = 20 kg 1 dog = 17kg

so we get the weight of each now we r going to sum them 1 rat + 1 cat + 1 dog = x

1 rat + 1 cat + 1 dog = x 3 kg + 7 kg + 17 kg = x

1 rat + 1 cat + 1 dog = x 3 kg + 7 kg + 17 kg = x 27 kg = x ..... is the mass of 3 of them

As we have proved that the series cosh(f(n)) ne1 diverges regardless of the rule for f, and that the series f(n) 2n³ - 1 converges regardless of the rule for f.

The comparison test states that if the terms of a series can be bounded below by a divergent series, then the given series also diverges.

In this case, we can bound the terms of cosh(f(n)) below by the series eⁿ. To see why, note that cosh(x) >= 1 for all x > 0. Thus, we have cosh(f(n)) >= 1 for all n. On the other hand, we know that e^x > 1 for all x > 0. Therefore, we have eⁿ > 1 for all n.

Since eⁿ diverges by the assumption that f satisfies O<f(), the comparison test tells us that cosh(f(n)) ne1 also diverges. Thus, the series cosh(f(n)) ne1 diverges regardless of the rule for f.

Moving on to the second part of the question, we are asked to show that the series ( f(n) 2n3 - 1 converges regardless of the rule for f. Again, we can use the comparison test to show convergence.

We can bound the terms of the given series by the series 1/n². To see why, note that for all n > 1, we have f(n) > 0 since the domain of f is restricted to R+. Thus, we have f(n)² < f(n) 2n³ - 1. Dividing both sides by n⁶, we get f(n)²/n⁶ < ( f(n) 2n³ - 1)/n⁶.

Now, note that the series 1/n² converges by the p-test (which states that the series 1/nᵃ converges if p > 1).

Therefore, by the comparison test, the series ( f(n) 2n³ - 1 also converges regardless of the rule for f.

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Complete Question:

Let f: R+R be a function that satisfies O<f() So for all x > 0. (a) Show that the series cosh(f(n)) ne1 diverges regardless of the rule for f. (b) Show that the series ( f(n) 2n3 - 1 converges regardless of the rule for f.

Sea "l" la medida del largo del terreno y "a" la medida del ancho del terreno.

De acuerdo con el problema, el ancho es la mitad del largo, es decir, a = l/2.

El perímetro de un rectángulo se calcula sumando las longitudes de sus cuatro lados, por lo que en este caso:

Perímetro = 2l + 2a = 2l + 2(l/2) = 3l

Sabemos que el perímetro es de 105 metros, entonces:

3l = 105

l = 105/3 = 35

Por lo tanto, el largo del terreno es 35 metros. Y, como el ancho es la mitad del largo, entonces:

a = l/2 = 35/2 = 17.5

Por lo tanto, el ancho del terreno es de 17.5 metros.

En resumen, las medidas del terreno son 35 metros de largo y 17.5 metros de ancho.

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The volume of the sample of the metal is 30 cubic centimeters.

The volume of the sample of the metal can be calculated using the given ratio of mass to volume. Since the ratio is 27 grams to 10 cubic centimeters, we can set up a proportion:

27 grams / 10 cubic centimeters = 81 grams / x cubic centimeters

Solving for x, we get:

x = (81 grams x 10 cubic centimeters) / 27 grams
x = 30 cubic centimeters


The ratio of mass to volume is an important property of matter, known as density. It describes how tightly packed the particles in a substance are. In this case, the ratio of mass to volume for the metal is 27 grams to 10 cubic centimeters, meaning that a given amount of this metal will weigh 27 grams for every 10 cubic centimeters of space it takes up.

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Answer:

  C. R+S+T = 201°

Step-by-step explanation:

You want to know which of the offered angle relations is true regarding quadrilateral RSTU.

The sum of angles in a quadrilateral is 360°. You use this fact to find angle T. Then you can compute the various differences to see which one matches the answer choices.

  T = 360° -R -S -U = 55°

In the attached calculator display, we have done exactly that. We find ...

  T -R = 38° . . . . A is false

  S -T = 74° . . . . B is false

  R +S +T = 201° . . . . C is TRUE

  R +T +U = 231° . . . . D is false

Answer:

To answer your question, we need to use some properties of rectangles and triangles.

A rectangle has four right angles, so angle R = angle S = angle T = angle U = 90 degrees.

The sum of the angles in a triangle is 180 degrees, so we can find the values of a, b, c, d, e, and f by using this property. For example, a + b + angle S = 180, so a + b = 90. Similarly, c + d = 90, e + f = 90, and f + g + angle U = 180, so f + g = 30.

Now we can evaluate each statement and see which one is true.

A) The difference between the measures of LT and LR is 4°. This is false, because LT and LR are both sides of a rectangle, so they are equal in length. The difference between them is zero, not four.

B) The difference between the measures of 2S and LT is 95°. This is false, because 2S is an angle and LT is a length. They have different units and cannot be compared or subtracted.

C) The sum of the measures of LR, 2S, and LT is 201°. This is false, because LR and LT are lengths and 2S is an angle. They have different units and cannot be added together.

D) The sum of the measures of LR, LT, and ZU is 193°. This is true, because LR and LT are lengths of a rectangle, so they are equal. ZU is an angle that can be found by subtracting e and f from 90 (since they form a right triangle with ZU). So ZU = 90 - e - f = 90 - (90 - c - d) - (90 - a - b) = a + b + c + d - 90. We know that a + b = c + d = 90, so ZU = 90 - 90 = 0.

Therefore, the sum of LR, LT, and ZU is LR + LT + 0 = 2LR = 2(17) = 34 degrees.

The correct answer is D.

Step-by-step explanation:

I hope that would help!!

Can I have Brainliest please?

Have a nice day

The price of the advertisement that originally costs $110 after a 15% markdown is $ 83.5.

Original price = $110

Markdown percent = 15%

Reduce in price = 15% of 110

To calculate this, we divide the percent by 100 and multiply it by the number given. Like in the above case, we divide 15 by 100 to get 0.15 and multiply it by 110.

= 0.15 * 110 = 16.5

Price after markdown = original price - reduction in price

= 110 - 16.5

= $ 83.5

Thus, the price for a quarter-page ad of the school yearbook which used to be $110 has come down to $83.5 after a 15 % markdown to attract more sponsors.

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To estimate your paycheck, a good percentage to multiply your earning by would be 0.75 or 75%. When calculating your paycheck, it's important to account for taxes, deductions, and other withholdings that may be taken out of your gross pay.  

This accounts for taxes, deductions, and other withholdings that are typically taken out of your paycheck before you receive your net pay. For example, if you earn $1,000 per pay period, multiplying by 0.75 would give you an estimated net pay of $750. However, keep in mind that this is just an estimate and your actual net pay may vary depending on your specific tax situation and other factors.

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The values of given conditions are:

1. median=70

2. Lower quartile=40

3. Upper quartile=87

4. Range=170

5. IQR=47

6. Lower outlier threshold=-20.5

7. Upper outlier threshold=160.5

In statistics, the median is the value separating the higher half from the lower half of a dataset. In other words, it is the middle value of a dataset when it is ordered in ascending or descending order.

Here,

To find the median time, we need to arrange the data in order from least to greatest and find the middle value.

10, 15, 30, 35, 40, 40, 45, 55, 60, 62, 64, 64, 66, 68, 70, 70, 72, 75, 75, 80, 82, 84, 90, 90, 105, 110, 120, 180

There are 28 values in the data set, so the median is the average of the 14th and 15th values:

Median = (70 + 70)/2

= 70

To find the lower quartile, we need to find the median of the lower half of the data set:

10, 15, 30, 35, 40, 40, 45, 55, 60, 62, 64, 64, 66, 68

There are 14 values in the lower half, so the lower quartile is the median of these values:

Lower quartile = (40 + 40)/2

= 40

To find the upper quartile, we need to find the median of the upper half of the data set:

72, 75, 75, 80, 82, 84, 90, 90, 105, 110, 120, 180

There are 14 values in the upper half, so the upper quartile is the median of these values:

Upper quartile = (84 + 90)/2

= 87

To find the range, we subtract the smallest value from the largest value:

Range = 180 - 10

= 170

To determine if there are any outliers in the data set, we need to calculate the interquartile range (IQR):

IQR = Upper quartile - Lower quartile

= 87 - 40

= 47

Any value that is more than 1.5 times the IQR below the lower quartile or above the upper quartile is considered an outlier.

Lower outlier threshold = Lower quartile - 1.5IQR

= 40 - 1.547

= -20.5

Upper outlier threshold = Upper quartile + 1.5IQR

= 87 + 1.547

= 160.5

To draw a box-and-whisker plot, we need to plot a box from the lower quartile to the upper quartile, with a line inside the box representing the median. We then draw whiskers extending from the box to the smallest and largest values that are not outliers. The box extends from 40 to 87, with a line at 70 representing the median. The whisker on the left extends to the smallest non-outlier value of 10, and the whisker on the right extends to the largest non-outlier value of 120.

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Answer: Nori

Step-by-step explanation:

2% of 5000 = 100

1% of 1000 = 10

Step-by-step explanation:

Pick any of the two points...I'll use the first two

calculate slope:     m  = ( y1-y2) / (x1-x2) =  (-14 - -5) / (-2 -1) = -9/-3 = 3

   equation of a line in slope intercept form is y = mx+ b

      so now you have    y = 3x + b  

             sub in any of the x,y  points given (8,16)  to calculate 'b'

                             16 = 3 (8) + b

                                  b = -8

so your first line is     y = 3x - 8

In a similar fashion, for the second one   m = - 5/8    and b = 2

           y = -5/8 x + 2

The value of y that satisfies the equation [tex]\int x^5 dx = \int x^y dx[/tex] is y = -1.

We know that the indefinite integral of x^5 dx is (1/6) x^6 + C, where C is

the constant of integration. Therefore:

[tex]\int x^5 dx = (1/6) x^6 + C[/tex]

We want to find y such that [tex]\int x^5 dx = \int x^y dx[/tex]. Using the power rule of integration, the indefinite integral of [tex]x^y[/tex] dx is [tex](1/(y+1)) x^{(y+1)} + C[/tex], where C is the constant of integration. Therefore:

[tex]\int x^y dx = (1/(y+1)) x^{(y+1)} + C[/tex]

For these two integrals to be equal, we need:

[tex](1/6) x^6 + C = (1/(y+1)) x^{(y+1) } + C[/tex]

Subtracting C from both sides, we get:

[tex](1/6) x^6 = (1/(y+1)) x^{(y+1)}[/tex]

Multiplying both sides by (y+1), we get:

[tex](1/6) x^6 (y+1) = x^{(y+1)}[/tex]

Now, we can equate the powers of x on both sides:

[tex]x^6 (y+1) = x^{(y+1)}[/tex]

Using the fact that[tex]x^a \times x^b = x^{(a+b)}[/tex], we can simplify the left-hand side:

[tex]x^(6(y+1)) = x^{(y+1)}[/tex]

Now, we can equate the exponents on both sides:

6(y+1) = y+1

Simplifying, we get:

6y + 6 = y + 1

5y = -5

y = -1

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Step-by-step explanation:

use Pythagorean theorem to find the height

c = 38 ft

a = 14 ft

a² + b² = c²

(14)² + b² = (38)²

b² = 1444 - 196

b² = 1248

b = √1248

b = √16 × 78

b = 4√78 feet

1.The universal set is defined as {Sarah, Allen, Kara, Todd, Joseph, Lydia, Matt, Caleb, Britney, Ethan}.

2.Caleb and Britney are included in both subsets since they like both football and basketball.

1. The universal set (U) consists of all the students in Mrs. Austin's class. In this case, U = {Sarah, Allen, Kara, Todd, Joseph, Lydia, Matt, Caleb, Britney, Ethan}.

2. The two subsets are:
 a) The set of students who like football (F) = {Sarah, Allen, Kara, Todd, Caleb, Britney}
 b) The set of students who like basketball (B) = {Joseph, Lydia, Matt, Caleb, Britney}

Caleb and Britney are included in both subsets since they like both football and basketball.

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