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aalyn [17]
3 years ago
12

An employee put $5,000 in a retirement account that offers 9% interest compounded annually. The employee makes no additional dep

osits or withdrawals. How much interest will the employee have earned at the end of 5 years. Make sure to round your answer to the hundredths.
Mathematics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

$2,693.12 will the employee have eared at the end of 5 years.

Step-by-step explanation:

The formula of compound interest:

A=P(1+r)^n

I = A-P

A= The amount in the account after n years

P= Principal

r= rate of interest annually

n= time in years.

An employee put $5,000 in retirement account that offer 9% interest compounded annually.

Here P=$5,000, r=9%=0.09, n= 5 years

A=5,000(1+0.09)^5

   =$7,693.12

The interest = A-P

                   =$(7,693.12-5,000)

                   =$2,693.12

$2,693.12 will the employee have eared at the end of 5 years.

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Answer:

Step-by-step explanation:

Let us first generate the frequency table from the information given:

Hurricane number(X)       Frequency(f)                f(X)

6                                            9                                   54

8                                            13                                  104

12                                           16                                  192

14                                            12                                168

Total                                     ∑(f) = 50                          ∑f(x) =518

In order to determine the last frequency (the remaining years), we will add the other frequencies and subtract the answer from 50, which is the total frequency (50 years). This is done as follows:

Let the last frequency be f

9 + 3 + 16 + f = 50

38 + f = 50

f = 50 - 38 = 12

Now, calculating mean:

\bar {X} = \frac{\sum f(x)}{\sum(f)} \\\\\bar {X} = \frac{518}{50} \\\\\bar {X} = 10.36

Therefore mean number of hurricanes = 10.4 (to one decimal place)

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Step-by-step explanation:

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Answer:

1) Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:  

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

\bar X=640 represent the sample mean

\sigma=150 represent the population standard deviation

n=40 sample size  

\mu_o =500 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

Step 2: Calculate the statistic

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

We can replace in formula (1) the info given like this:  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

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