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2 years ago

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in 1950, the population of a city was about 800,000 people. over the next 50 years , the population increased to 1.3 million. fi

nd the percent of increase

1 answer:

nordsb [41]2 years ago

5 0

Subtract 1.3 million from 800,000 and you get ur answer

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If f(1) = 10, what is f(3)?

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Answer:

30

Step-by-step explanation:

multiply each f by the same root numeral

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I was shot in My head with an ar15 containing radar identity atom shortly after receiving another Ph.d. in neuropsychology ident

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2 years ago

The area of Terry's room at the old house was 96 ft.2. The area of Terry's room at the new house is 108 ft.2

Finger [1]

Answer:

12.5%

Step-by-step explanation:

First we need to calculate the change in the area of Terry's room this is

108 ft2 - 96 ft2 = 12 ft2

Then we use a direct proportion in order to find the percentage

96 ft2 ----- 100%

12 ft2 ----- p?

Solvin for p we have that

$p=\frac{12\times 100}{96}$

p=12.5

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3 years ago

Read 2 more answers

What is the product of 8/6 and 5/10 in simplest radical form?

Gre4nikov [31]

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2 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

2 years ago