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maw [93]
1 year ago
9

Evaluate b-(-1/8)+c where b=2 and c=-7/4

Mathematics
1 answer:
bogdanovich [222]1 year ago
3 0

Answer: \Large\boxed{\dfrac{3}{8} }

Step-by-step explanation:

<u>Given information</u>

b=2

c=-\dfrac{7}{4}

<u>Given expression </u>

b-(-\dfrac{1}{8}) +c

<u>Substitute values into the expression</u>

=(2)-(-\dfrac{1}{8}) +(-\dfrac{7}{4} )

<u>Convert the fractions into the common denominator</u>

<em>Least Common Multiple (LCM) of 8 and 4 = 8</em>

=(2)-(-\dfrac{1}{8}) +(-\dfrac{7\times2}{4\times2} )

=(2)-(-\dfrac{1}{8}) +(-\dfrac{14}{8} )

<u>Simplify the parenthesis</u>

=2+\dfrac{1}{8} -\dfrac{14}{8}

<u>Simplify by addition</u>

=\dfrac{16}{8} +\dfrac{1}{8} -\dfrac{14}{8}

=\dfrac{17}{8}  -\dfrac{14}{8}

<u>Simplify by subtraction</u>

\Large\boxed{=\dfrac{3}{8} }

Hope this helps!! :)

Please let me know if you have any questions

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In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
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Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

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Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

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(c)

Compute the 95% confidence interval for population proportion as follows:

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Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

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