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snow_lady [41]
3 years ago
12

How would the shape of the distribution change if the salesman decides to also deal in cars priced under $5,000 and in cars pric

ed from $45,000 to $50,000 and projects sales of 200 cars in each category?

Mathematics
1 answer:
Anna35 [415]3 years ago
5 0
This should help you answer your problem 

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Divide 61) 213.5 what is the answer
IrinaK [193]

Answer:

3.5

Step-by-step explanation:

5 0
3 years ago
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Marsha sees an airplane flying overhead. The spot on the ground that is directly under the plane is 3,600 feet from Marsha's hou
Rudiy27

The elevation of the plane is 3600 ft

Let

  • h = elevation of plane,
  • d = distance from Marsha's house to spot on ground below plane = 3600 feet and
  • Ф = angle of elevation of plane = 45°

From the triangle, we have the trigonometric ratio

tanФ = h/D

<h3>Elevation of the plane</h3>

Making h subject of the formula, we have

h = DtanФ

substituting the values of the variables into the equation, we have

h = DtanФ

h = 3600 ft × tan45°

h = 3600 ft × 1

h = 3600 ft

So, the elevation of the plane is 3600 ft

Learn more about elevation of the plane here:

brainly.com/question/26380084

7 0
2 years ago
Graph help fast!!!!!!!!!!1111
ki77a [65]

Answer:

(0,0)

(1,3)

(2,6)

(3,9)

(-1,-3)

(-2,-6)

(-3,-9)

Step-by-step explanation:

The equation of the graph is y=3x

5 0
2 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
The base of a prism are right triangles with leg lengths of 7 inches and 5 inches. the prism height is 12 inches. what is the vo
Wewaii [24]
The volume formula is length x width x height. 7x5x12=420 cubic inches
4 0
2 years ago
Read 2 more answers
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