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ipn [44]
3 years ago
5

The population of big cats in Africa is increasing at a rate of 5% per year. At the

Mathematics
1 answer:
sesenic [268]3 years ago
6 0

Answer:

a) 10,500 big cats

b) 13,401 big cats

c) 33 years

Step-by-step explanation:

Hi, to answer this question we have to apply an exponential growth function:  

A = P (1 + r) t  

Where:  

p = original population  

r = growing rate (decimal form; 5/100= 0.05)

t= years  

A = population after t years  

Replacing with the values given:  

a) At 2005, 1 year passed since 2004 (t=1)

A = 10,000 (1+ 0.05)^1

A = 10,500 big cats

b)

At 2010, 6 years passed since 2004 (2010-2004= 6)

A = 10,000 (1+ 0.05)^6

A = 13,401 big cats

c)

50,000 < 10,000 (1+ 0.05)^t

Solving for t:

50,000/10,000 < 1.05^t

5 < 1.05^t

log 5 < log 1.05^t

log 5 < t ( log 1.05)

log 5 / log 1.05 < t

32.9 years < t

33 years  = t

Feel free to ask for more if needed or if you did not understand something.  

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The friend caught the ball at 2 feet.

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3 years ago
At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500
kaheart [24]

Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}

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i.e. PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}

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Hello from MrBillDoesMath!

Answer:

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See attached doc and ask questions!

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