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3 years ago

Analia wants to know which school has higher athletic achievement relative to the budget per student

Mathematics

1 answer:

user100 [1]3 years ago

7 0

Answer:

SAT average divided by budget per student

Graduation rate divided by budget per student

No. The definitions have opposite results.

Step-by-step explanation:

DID IT ON KHAN and

The key understanding here is that there isn't a single way to define "academic achievement." Graduation rate can be a good measure of that. Average SAT is also a good measure. The rate of students who are accepted into college can also be a good measure, but we are not given information about that.

Now let's go over the options and see which of them might be used to describe "academic achievements relative to the budget per student."

Hint #22 / 4

Definition Analysis Appropriate?

SAT average divided by budget per student "SAT average" is a measure of academic achievement. Dividing this by the budget per student gives us a good definition.

Yes

Yesstart color #1fab54, start text, Y, e, s, end text, end color #1fab54

Graduation rate divided by budget per student "Graduation rate" is a measure of academic achievement. Dividing this by the budget per student gives us a good definition.

Yes

Yesstart color #1fab54, start text, Y, e, s, end text, end color #1fab54

Number of sports medals won divided by budget per student "Number of sports medals won" isn't a measure of academic achievement.

Nostart color #e84d39, start text, N, o, end text, end color #e84d39

Graduation rate divided by number of sports medals won "Graduation rate" is a measure of academic achievement, but it's divided by the number of sports medals won, and not by the budget per student.

Nostart color #e84d39, start text, N, o, end text, end color #e84d39

Hint #33 / 4

We are left with two definitions for academic achievements relative to the budget per student:

SAT average divided by budget per student

Graduation rate divided by budget per student

Let's calculate the academic achievements relative to the budget per student for each school according to each definition:

School Definition #1 Definition #2

≈

0.114

student points

dollar

≈0.114

dollar

student points

approximately equals, 0, point, 114, start fraction, start text, s, t, u, d, e, n, t, space, p, o, i, n, t, s, end text, divided by, start text, d, o, l, l, a, r, end text, end fraction

≈

0.0082

student percents

dollar

≈0.0082

dollar

student percents

approximately equals, 0, point, 0082, start fraction, start text, s, t, u, d, e, n, t, space, p, e, r, c, e, n, t, s, end text, divided by, start text, d, o, l, l, a, r, end text, end fraction

0.105

student points

dollar

0.105

dollar

student points

0, point, 105, start fraction, start text, s, t, u, d, e, n, t, space, p, o, i, n, t, s, end text, divided by, start text, d, o, l, l, a, r, end text, end fraction

0.009

student percents

dollar

0.009

dollar

student percents

0, point, 009, start fraction, start text, s, t, u, d, e, n, t, space, p, e, r, c, e, n, t, s, end text, divided by, start text, d, o, l, l, a, r, end text, end fraction

Which school has higher academic achievements relative to the budget per student? School A School B

[Show me the calculations used for this table.]

Hint #44 / 4

In conclusion:

1) These are the two definitions Analia came up with to define the quantity:

SAT average divided by budget per student

Graduation rate divided by budget per student

2) The definitions have opposite results.

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Step-by-step explanation:

I could <u><em>only</em></u> work on questions 9, 10, 11, 12 in accordance with Brainly's rules. Nevertheless, the techniques demonstrated in this post applies to all of the given problems in your worksheet.

<h2><u>Definitions:</u></h2>

The given set of graphs are examples of absolute value functions. The <u>general form</u> of absolute value functions is: y = a|x – h| + k, where:

|a| = determines the vertical stretch or compression factor (wideness or narrowness of the graph).

(h, k) = vertex of the function

x = h represents the axis of symmetry.

<h2><u>Solutions:</u></h2><h3>Question 9) ⇒ Vertex: (-4, 2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (0, 5) and substitute into the general form (equation):

y = a|x – h| + k

5 = a| 0 - (-4)| + 2

5 = a| 0 + 4 | + 2

5 = a|4| + 2

5 = 4a + 2

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5 - 2 = 4a + 2 - 2

3 = 4a

Divide both sides by 4 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{3}{4}\:=\:\frac{4a}{4}}$

a = ¾

Therefore, given the value of a = ¾, and the vertex, (-4, 2), then the equation of the absolute value function is:

<u>Equation</u>: y = ¾|x + 4| + 2

<h3>Question 10) ⇒ Vertex: (0, -3)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -2) and substitute into the general form (equation):

y = a|x – h| + k

-2 = a|4 - 0| -3

-2 = a|4| - 3

-2 = 4a - 3

Add 3 to both sides:

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Divide both sides by 4 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{1}{4}\:=\:\frac{4a}{4}}$

a = ¼

Therefore, given the value of a = ¼, and the vertex, (0, -3), then the equation of the absolute value function is:

<u>Equation</u>: y = ¼|x - 0| - 3

<h3>Question 11) ⇒ Vertex: (3, 1)</h3>

<u>Solve for a:</u>

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y = a|x – h| + k

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-3 - 1 = a + 1 - 1

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<u>Equation</u>: y = -4|x - 3| + 1

<h3>Question 12) ⇒ Vertex: (-2, -2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (-4, 0) and substitute into the general form (equation):

y = a|x – h| + k

0 = a|-4 - (-2)| - 2

0 = a|-4 + 2| - 2

0 = a|-2| - 2

0 = 2a - 2

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0 + 2 = 2a - 2 + 2

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Divide both sides by 2 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{2}{2}\:=\:\frac{2a}{2}}$

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