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3 years ago

Find the commission on a $750 sale if the commission is 24%

Mathematics

1 answer:

Savatey [412]3 years ago

6 0

Answer:

The commission is $180

Step-by-step explanation:

To find the commission take the total (750) and multiply it by the percentage of commission (.24).

When you do this, you will get $180.

Hope this helps! :)

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Solve using substitution for x and y 5x+y=16 <br> y=2-3x

valentina_108 [34]

Answer:

X = 7 Y = -19

Step-by-step explanation:

If you want to solve this equation by substitution then you have to put the second equation in, so it will look something like this

5x + 1(2-3x)=16

Apply distributive property to solve.

Once you find x, plug it in to find y

To double check your work, you can plug the coordinated values back in

-19 = 2 - 3(7)

-19 = 2-21

This is true, so our solution is correct.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Lydia collects shells. She has 24 sea snail shes

Wittaler [7]

Answer:

Step-by-step explanation:

the greatest common factor is 8 so you know you have to have 8 rows, divide all of those number by 8 and add the answers

8 0

4 years ago

Read 2 more answers

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.