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hram777 [196]
3 years ago
6

2) Marion deposited $12,000 into her saving account for 10 years with simple annual interest rate of 5%. Cameron deposited $12,0

00 into his saving account with annual compound interest rate of 4% for 10 years. Which account will have more money after 10 years and by how much.
Mathematics
1 answer:
morpeh [17]3 years ago
6 0

Answer:

Marion’s account will have $237 more at the end of 10 years

Step-by-step explanation:

Firstly, we calculate the amount that will be in Marion’s account after 10 years.

To calculate this, we use the formula for simple interest

I = PRT/100

where I is the interest accrued for the period of years

P is the amount deposited = $12,000

R is the rate = 5%

T is the time which is 10 years

Plugging these values into the equation

I = (12,000 * 5 * 10)/100 = $6,000

The amount after 10 years is thus the sum of the amount deposited and the interest accured = $12,000 + $6,000 = $18,000

Now for Cameron, we use the compound interest formula

A = P(1+r/n)^nt

Where A is the amount in the account after the number of years

P is the amount deposited = $12,000

r is the interest rate = 4% = 4/100 = 0.04

n is the number of times per year the interest is compounded. Since it is annually, n = 1

t is the time which is 10 years

We plug these values and we have;

A = 12,000(1 + 0.04/1)^(1 * 10)

A = 12,000 (1.04)^10

A = $17,763 ( to the nearest whole dollars)

Since 18,000 is greater than 17,763, the amount in Marion’s account will be greater at an amount of (18,000 - 17,763) = $237

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Answer:

95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                              P.Q. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean time = 5.15 years

            \sigma = sample standard deviation = 1.68 years

            n = sample of college graduates = 4400

            \mu = population mean time

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                               level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                              = [ 5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } } , 5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } } ]

                                             = [5.10 , 5.20]

Therefore, 95​% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0
3 years ago
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