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Snowcat [4.5K]
2 years ago
7

The day you were born your Grandparents started to save money for your future education. They decided to deposit $3,000 at the e

nd of each year, for eighteen years, in a savings account that pays 7.5% per year, compounded annually.
On your 19th birthday you get admission into university for a four-year degree course. You calculate that you require approximately $35,000, for your fees, books and living expenses for each year. Assuming you keep the money in the bank accruing the same interest, and withdraw $ 35,000, at the beginning of each year.

a) Calculate the total value of this investment at the end of the term?

b) Determine the total interest earned?

Will your inheritance last you for the full tenure of your four years at the university?
​
Mathematics
1 answer:
cestrela7 [59]2 years ago
5 0

The total value of the investment is $107,032.16.

The total interest earned is $53,032.16

The inheritance would last the tenure of the university.

<h3>What is the total value of the investment?</h3>

The formula that can be used to determine the future value of the 18-year annuity is: annuity factor x amount deposited yearly

Annuity factor = {[(1+r)^n] - 1} / r

Where:

  • r = interest rate
  • n = number of years

$3000 x [(1.075^18 - 1) / 0.075] = $107,032.16

Amount deposited in the course of 18 years = (3000 x 18) = 54,000

Interest earned = 107,032.16 - 54,000 = $53,032.16

To learn more about annuities, please check: brainly.com/question/24108530

#SPJ1

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(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X_{M}= 69.8 represent the mean for the age of grandmother

\bar X_{F}= 71.6 represent the mean for the age of grandfather

s_{M}= 8.38 represent the sample deviation for the age of grandmother

s_{F}= 6.65 represent the sample deviation for the age of grandfather

n_M = n_F= 10

Solution to the problem

For this case the confidence interval is given by:

(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n_M +n_F-2=10+10-2=18

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that t_{\alpha/2}=2.101

And replacing we got:

(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

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