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Rina8888 [55]
3 years ago
5

Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors

. A sample of 300 students taken from this year's student body showed the following number of students in each classification.
Freshmen 83
Sophomores 68
Juniors 85
Seniors 64
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year.

Mathematics
1 answer:
Natalija [7]3 years ago
6 0

Answer:

There has been no change in the classifications between the last school year and this school year.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H</em>₀: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}

The expected values are computed using the formula:

E_{i}=p_{i}\times N

Here, <em>N</em> = 300

Use Excel to compute the values.

The test statistic value is:

\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}=1.662

The test statistic value is, 1.662.

The degrees of freedom of the test is:

<em>n</em> - 1 = 4 - 1 = 3

The significance level is, <em>α</em> = 0.05.

Compute the p-value of the test as follows:

<em>p</em>-value = 0.6454

*Use a Chi-square table.

p-value = 0.6454 > <em>α</em> = 0.05.

So, the null hypothesis will not be rejected at 5% significance level.

Thus, concluding that there has been no change in the classifications between the last school year and this school year.

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Answer:

C. x^2 -4

Step-by-step explanation:

What you do is subtract the functions as said by

(f-g)(x).

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Answer:

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Step-by-step explanation:

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The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate
soldi70 [24.7K]

Answer:

a) V = 33510.322\,in^{3}, b) A_{s} = 5026.548\,in^{2}, c) \% V = 0.450\,\%, \%A_{s} = 0.300\,\%.

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

V = \frac{4}{3}\pi \cdot r^{3}

A_{s} = 4\pi \cdot r^{2}

a) The volume of the sphere is:

V = \frac{4}{3}\pi \cdot (20\,in)^{3}

V = 33510.322\,in^{3}

b) The surface area of the sphere is:

A_{s} = 4\pi \cdot (20\,in)^{2}

A_{s} = 5026.548\,in^{2}

c) The total differentials for volume and surface area of the sphere are, respectively:

\Delta V = 4\pi\cdot r^{2}\,\Delta r

\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)

\Delta V = 150.796\,in^{3}

\Delta A_{s} = 8\pi\cdot r \,\Delta r

\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)

\Delta A_{s} = 15.080\,in^{2}

Relative errors are presented hereafter:

\%V = \frac{\Delta V}{V}\times 100\%

\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%

\% V = 0.450\,\%

\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%

\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%

\%A_{s} = 0.300\,\%

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3 years ago
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Paladinen [302]

The answers are A and D. Hope this helps!

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