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spayn [35]
4 years ago
11

A person inventes 4500 dollars in a bank. The bank pays 4.75% interest compounded semi-annually. To the nearest tenth of a year,

how long mist the person leave the money in the bank until it reaches 5900 dollars
Mathematics
1 answer:
mars1129 [50]4 years ago
5 0

Answer:

It should take 5.8 years to reach his goal.

Step-by-step explanation:

In order to solve this question we must use the compounded interest's formula shown below:

M = C*(1 + r/n)^(t*n)

Where M is the final amount, C is the initial amount, r is the interest rate, n is the compound rate and t is the time elapsed in years. Applying the data from the problem we have:

5900 = 4500*(1 + 0.0475/2)^(2*t)

5900 = 4500*(1 + 0.02375)^(2*t)

5900 = 4500*(1.02375)^(2*t)

4500*(1.02375)^(2*t) = 5900

(1.02375)^(2*t) = 5900/4500

(1.02375)^(2*t) = 59/45

log[1.02375^(2*t)] = log(59/45)

2*t*log(1.02375) = log(59/45)

t = log(59/45)/[2*log(1.02375)] = 5.77008

It should take 5.8 years to reach his goal.

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Answer:

a) I(95,50) = 73.19 degrees

b) I_{T}(95,50) = -7.73

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

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(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is I_{T}(T,H). So

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H

I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73

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