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Setler [38]
3 years ago
9

Workers on the assembly line produce 4x + 6 boats each day. Which expression shows how many boats they produced in 12 days?

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
4 0
The answer would be A. If 4x + 6 is the number of boats made in 1 day, then multiplying that by 12 would give us the number of boats made in 12 days so the equation would look like this:
12(4x + 6)
podryga [215]3 years ago
4 0

The answer would be A)12(4x+6),because there are twelve days and your multiplying 12 by the number of boats which is 4x+6 so it would be 12(4x + 6).

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8n-13=13-8n can someone answer with steps, please?
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Your answer is 1.625

Hope I helped :)

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(unless the other person does a better job)

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4 0
3 years ago
26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

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