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3 years ago

Unit rate for 150 students in 3 classes.

Mathematics

2 answers:

damaskus [11]3 years ago

7 0

Answer:

Step-by-step explanation:

150 students/3 classes= 50 students per class

balu736 [363]3 years ago

6 0

50 students per class 150/3 = 50

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Solve the following and show all work: √8x+1=x+2

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x= 3, 1

Step-by-step explanation:

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3 years ago

Edger Anderson earns $300 A week plus a 15 percent commission only on sales he makes after his first $1000 In sales.if Mr.Anders

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525 is what i got as my answer good luck

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4 years ago

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Consider the following data to determine whether there is a significant difference between the values of Group 1 and Group 2. Le

cestrela7 [59]

Using the t-distribution, it is found that since the p-value of the test is 0.72 > 0.05, there is not significant difference between the values of each group.

At the null hypothesis, we test if the groups have the same values, that is, the subtraction of their means is 0:

$H_0: \mu_2 - \mu_1 = 0$

At the alternative hypothesis, we test if the groups have different values, that is, the subtraction of their means is different of 0:

$H_1: \mu_2 - \mu_1 \neq 0$

First, we have to find the mean and the standard deviation for both samples, hence, using a calculator:

$\mu_1 = 18.75, \sigma_1 = 6.2507, \mu_2 = 19.75, \sigma_2 = 4.2678$

The standard errors are:

$s_1 = \frac{\sigma_1}{\sqrt{n_1}} = \frac{6.2507}{\sqrt{8}} = 2.21$

$s_2 = \frac{\sigma_2}{\sqrt{n_2}} = \frac{4.2678}{\sqrt{8}} = 1.5089$

For the distribution of differences, we have that:

$\overline{x} = \mu_2 - \mu_1 = 19.75 - 18.75 = 1$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.21^2 + 1.5089^2} = 2.676$

We have the standard deviation for the samples, hence, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu$ is the value tested at the null hypothesis, which is 0 for this problem. Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{1 - 0}{2.676}$

$t = 0.37$

To find the p-value, we have a two-tailed test, as we test if the means are different from a value, with t = 0.37 and 8 + 8 - 2 = 14 df.

Using a t-distribution calculator, this p-value is of 0.72.

Since the p-value of the test is 0.72 > 0.05, there is not significant difference between the values of each group.

A similar problem is given at brainly.com/question/15682365

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