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1 year ago

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Suppose in the next year, 2007, College D's expenses and enrollment remain about the same, but in addition to their current reve

nues, they receive an additional $50,000,000 grant. This would allow them to reduce average tuition by how much?

1 answer:

Olin [163]1 year ago

7 0

This question is incomplete, here is the complete question:

Suppose in the next year, 2007, College D's expenses and enrollment remain about the same, but in addition to their current revenues, they receive an additional $50,000,000 grant. This would allow them to reduce average tuition by how much?

A) $1388.89

B) $3571.43

C) $5555.56

D) $9500.00

E) $25888.89

number of students = 36,000

Answer: A) $

1388.89

Step-by-step explanation:

the college received additional grant which is $50,000,000

and the number of students is 36,000,

and we also know that expenses and enrollment remained the same.

So if we have more money (grants) and nothing changed (expenses remain the same)

dividing the grant by the number of students will show just how much the average tuition fee would be reduced

therefore R = G/n

R = 50,000,000 / 36000

R = 1,388.888 ≈ $1388.89

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The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr

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Answer:

(a) $Perimeter = 32.2\ cm$

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

$Area = 64.8$

Required

Calculate the perimeter.

Area is calculated as:

$Area = L * W$

Where

$L = Length$ and $W = Width$

Substitute 64.8 for Area

$64.8 = L * W$

Make L the subject:

$L = \frac{64.8}{W}$

Perimeter is calculated as:

$P = 2 * (L + W)$

Substitute 64.8/W for L

$P = 2 * (\frac{64.8}{W} + W)$

$P = \frac{129.6}{W} + 2W$

To solve further, we take the derivative of P and set it to 0, afterwards.

$dP = -\frac{129.6}{W^2} + 2$

Set to 0

$0 = -\frac{129.6}{W^2} + 2$

Collect Like Terms

$\frac{129.6}{W^2} = 2$

Cross Multiply:

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Divide through by 2

$W^2 = 64.8$

Take square roots

$W = \sqrt{64.8$

$W = 8.05$

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$Perimeter = 2 * (16.10)$

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Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

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$Tenths = \frac{1}{10}$

So, the expression becomes:

((1 Group of $10 * \frac{1}{10}$ ) and (1 group of $8 * \frac{1}{10}$ ))/(6 groups of $3 * \frac{1}{10}$ )

This gives:

((1 Group of $\frac{10}{10}$ ) and (1 group of $\frac{8}{10}$ ))/(6 groups of $\frac{3}{10}$ )

Group means product, so the expression becomes:

$\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

And, as used here means addition

$\frac{(1 * \frac{10}{10} + 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

Simplify:

$\frac{(1 * 1 + 1 * 0.80)}{6 *0.30}$

$\frac{(1 + 0.80)}{1.80}$

$\frac{1.80}{1.80}$

$= 1.00$

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Step-by-step explanation:

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