Shira's math test included a survey question asking how many hours students spent studying for the test. The scatter plot below shows the relationship between how many hours students spent studying and their score on the test. A line was fit to the data to model the relationship. Which of these linear equations best describes the given model? Choose 1 Choose 1 (Choice A) A \hat y=x+45 y ^ =x+45y, with, hat, on top, equals, x, plus, 45 (Choice B) B \hat y=10x+45 y ^ =10x+45y, with, hat, on top, equals, 10, x, plus, 45 (Choice C) C \hat y=-10x+45 y ^ =−10x+45y, with, hat, on top, equals, minus, 10, x, plus, 45 Based on this equation, estimate the score for a student that spent 555 hours studying.

Part 1)  y=10x+45

Part 2) The score is 95

Step-by-step explanation:

Part 1) Linear equation that best describes the given model

Let

x - number of hours students spent studying

y -  their score on the test

Looking at the line that was fit to the data to model the relationship

The slope is positive

The y-intercept is the point (0,45)

For x=1, y=55 ----> point (1,55)

Find the slope

The formula to calculate the slope between two points is equal to

substitute  the points (0,45) and (1,55)

Find the equation of the line in slope intercept form

we have

substitute

y=10x+45

Part 2)

Estimate the score for a student that spent 5 hours studying.  

For x=5 hours

substitute in the linear equation and solve for y

y=10(5)+45=95

y = 95

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

1) is right

step-by-step explanation: