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Olin [163]
3 years ago
7

Barry wants to make a drawling this is 1/2 the size of the original. If a tree in the original drawling is 11 inches tall and 7

inches wide, what will be the length and width of the tree in Barry's drawling?
Mathematics
1 answer:
avanturin [10]3 years ago
8 0

Answer: 5.5 inches, 3.5 inches

Step-by-step explanation:

Given

Barry wants to make a drawing i.e. half the size of original

The length and width of the original drawing are  11 inches and 7 inches respectively.

So, the intended drawing is obtained by multiplying the original dimensions by \frac{1}{2}

The length of barry's drawing is

\Rightarrow 11\times \frac{1}{2}\\\\\Rightarrow 5.5\ \text{inches}

The breadth of barry's drawing

\Rightarrow 7\times \frac{1}{2}\\\\\Rightarrow 3.5\ \text{inches}

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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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2 years ago
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