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horsena [70]
4 years ago
5

There were 900 students enrolled in a high school in 2009 and 1,500 students enrolled in the same high school in 2012. The stude

nt enrollment of the high school, P, has increased at a constant rate each year, t, since 2009.
A) Write the given enrollment numbers as a pair of points in the form (t, P).
B) Find the slope of the line passing through the pair of points from part A and explain the information the slope gives about the situation.
C) Write an equation that relates the high school's student enrollment, P, to the number of years since 2009, t.
D) Predict the high school's enrollment in 2017.
Mathematics
1 answer:
Volgvan4 years ago
4 0

Answer:

(A)

(t_0,P_0)=(0,900)

(t_1,P_1)=(3,1500)

(B)

m=200

(C)

P=200t+900

(D)

The high school's enrollment in 2017 is 2500

Step-by-step explanation:

Let's assume time starts since 2009

so, In 2009 , t=0

P=900

In 2012,

t=2012-2009=3

P=1500

(A)

So, we have points as

(t_0,P_0)=(0,900)

(t_1,P_1)=(3,1500)

(B)

we can use slope formula

m=\frac{P_2-P_2}{t_2-t_1}

we can plug values

m=\frac{1500-900}{3-0}

m=200

we know that

slope is rate of change of P with respect to time

so, slope means increase in population is 200 per year

(C)

we can use point slope form of line

y-y_1=m(x-x_1)

m=200

(t_0,P_0)=(0,900)

P-900=200(t-0)

P=200t+900

(D)

In 2017 ,

t=2017-2009=8

we can plug t=8

and we get

P=200(8)+900

P=2500

The high school's enrollment in 2017 is 2500

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Find attached the missing cash flow stream

Answer:

  • <u><em>Present value = $1,685,334 (rounded to the nearest whole)</em></u>

Explanation:

Since the <em>cash flow stream</em> is <em>uneven</em>, you must discount each stream individually and after you have discounted every stream you can add each preset value to find the net present value.

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               \text{Present Value} = \dfrac{\text{Future Value}}{(1+r)^n}\\\\\  Where:\\ \\ r = \text{Rate of return=interest rate}\\ \\ n = \text{Number of periods}

<u>1. Year 1:</u>

  • Future value = $250,000
  • r = 4%
  • n = 1

Present\text{ }value=\$250,000/(1+0.04)^1=\$240,384.62

<u>2. Year 2:</u>

  • Future value = $20,000
  • r = 4%
  • n = 2

Present\text{ }value=\$20,000/(1+0.04)^2=\$18,491.12

<u>3. Year 3:</u>

  • Future value = $330,000
  • r = 4%
  • n = 3

Present\text{ }value=\$330,000/(1+0.04)^3=\$293,368.80

<u>4. Year 4:</u>

  • Future value = $450,000
  • r = 4%
  • n = 4

Present\text{ }value=\$450,000/(1+0.04)^4=\$384,661.89

<u>5. Year 5:</u>

  • Future value = $550,000
  • r = 4%
  • n = 5

Present\text{ }value=\$550,000/(1+0.04)^5=\$452,059.91

<u>6. Year 6:</u>

  • Future value = $375,000
  • r = 4%
  • n = 6

Present\text{ }value=\$375,000/(1+0.04)^6=\$296,367.95

Total present value = $240,384.62 + $18,491.12 + $293,368.80 + $384,661.89 + $452,059.91 + $296,367.95 + $ 296,367.95

Total present value = $1,685,334.29 = $ 1,685,334

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3 years ago
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