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choli [55]
3 years ago
13

A student took a chemistry exam where the exam scores were mound-shaped with a mean score of 90 and a standard deviation of 64.

She also took a statistics exam where the scores were mound-shaped, the mean score was 70 and the standard deviation was 16. If the student's grades were 102 on the chemistry exam and 77 on the statistics exam, then: a. the student did relatively better on the chemistry exam than on the statistics exam, compared to the other students in each class. b. it is impossible to say which of the student's exam scores indicates the better performance. c. the student did relatively better on the statistics exam than on the chemistry exam, compared to the other students in the two classes. d. the student's scores on both exams are comparable, when accounting for the scores of the other students in the two classes. e. the student did relatively the same on both exams
Mathematics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

wow that school has alot of test don't it lol

Step-by-step explanation:

anyway what's the question

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first you have to find the triangle, then you can move to the actual question.

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7 0
3 years ago
Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =
marta [7]

Answer:

The integral for the arc of length is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

By using Simpon’s rule we get: 1.5355453

And using technology we get:  2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx

We need the derivative of the function:

f'(x)=\frac{1}{x}

And we need it squared:

[f'(x)]^2=\frac{1}{x^2}

Then the integral is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

Now, the Simposn’s rule with n=4 is:

\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)

In this problem:

a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}

So, the Simposn’s rule formula becomes:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then simplifying a bit:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then we just do those computations and we finally get the approximation via Simposn's rule:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%

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Answer:

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