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2 years ago

A sum of money is deposited in a bank which offers a simple interest rate of 0.325% per annum. At

Mathematics

1 answer:

nikitadnepr [17]2 years ago

3 0

Answer:

The sum of money deposited is approximately $22,021.74

Step-by-step explanation:

The given interest and amounts of the deposit are outlined as follows;

The simple interest per annum, R = 0.325%

The number of years the money is deposited, T = 4 years

The total amount received (Interest + Initial deposit), A = $50,650

We have;

I = P × R × T

Where;

P = The principal (initial amount deposited)

R = The (annual) interest rate = 0.325%

T = The time = 4 years

Therefore;

The total amount received, A = P + I

P + I = P + P × R × T = P × (1 + R × T)

∴ A = P + I = P × (1 + R × T)

P = A/(1 + R × T)

Plugging in the values, gives;

P = 50,650/(1 + 0.325 × 4) = 506,500/23 ≈ 22,021.74

The sum of money deposited, P = $22,021.74

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Step-by-step explanation:

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3 years ago

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horrorfan [7]

Answer:

Step-by-step explanation:

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2 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co

Ira Lisetskai [31]

Answer:

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

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Step-by-step explanation:

The first step is finding the confidence interval

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The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$ .

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$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

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3 years ago