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3241004551 [841]
3 years ago
11

Can u help with this?! One that solves best gets brainliest

Mathematics
1 answer:
lawyer [7]3 years ago
7 0

Answer:

2.) 3/8

3.) 3/8

4.) Couldn't read this one, too blurry

5.) 5/12

6.) 3/10

7.) 5/12

8.) 1/8

9.) 1/5

10.) 1/8

11.) 1/4

Step-by-step explanation:

So the thing about subtracting fractions is that you have to have the same denominator for both of them. (that's the number on the bottom.) So number 3 is 7/8 - 1/2. You're gonna wanna multiply 1/2 by four, for both the top and bottom number. This creates 4/8. Now you can subtract them! 7-4 = 3, and thus the answer is 3/8. You never, <em>ever</em> subtract the denominator: only the numerator.

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Creation of reproductive cells (gametes) requires...​
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A local company is concerned about the number of days missed by its employees due to ill ness. A Random Sample of 10 employees i
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Answer:

The incentive program does not cuts down on the number of days missed by employees.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

A paired <em>t</em>-test would be used to determine whether the incentive program cuts down on the number of days missed by employees.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: The incentive program does not cuts down on the number of days missed by employees, i.e. <em>d</em> ≥ 0.

<em>Hₐ</em>: The incentive program cuts down on the number of days missed by employees, i.e. <em>d</em> < 0.

From the information provided the data computed is as follows:

 n=10\\\bar d=-1.1\\SD_{d}=2.99

Compute the test statistic value as follows:

 t=\frac{\bar d}{SD_{d}/\sqrt{n}}

   =\frac{-1.1}{2.99/\sqrt{10}}\\\\=-1.16

The test statistic value is -1.16.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value of the test as follows:

p-value=P(t_{n-1}

*Use a t-table.

The p-value of the test is 0.1379.

The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.

Thus, there is not enough evidence to support the claim.

Conclusion:

The incentive program does not cuts down on the number of days missed by employees

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How would you describe the solution to this system of equations?
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Answer:

reffrents a number

Step-by-step explanation:

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3 years ago
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