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lana [24]
3 years ago
11

Jackson earned $60 and put 1 5 of those earnings into his savings account. How much did he put into his savings account? $

Mathematics
1 answer:
alex41 [277]3 years ago
5 0

Answer:

12 dollars

Step-by-step explanation:

60 divided by 5=12

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1.5 cookies and 0.5 brownies are needed for 1 plate. In 10 plates, you will need 15 cookies and 5 brownies.
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A student cost for last semester at her community college was $2900. She spent $464 of that on books. What percent of last semes
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Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test
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Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\  \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}

- Applying the formula, we have:

\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\  \\ t=\frac{5}{3.1752} \\  \\ t\approx1.5747 \end{gathered}

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

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2 years ago
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