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vovangra [49]
3 years ago
12

Solve the attachment...​

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx

Here 5 is a constant so it can come out . So that,

\longrightarrow \displaystyle I =  5 \int_0^1 x \sqrt{x}\ dx

Now we can write √x as ,

\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx

Simplify ,

\longrightarrow I =  5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx

By Power rule , the integral of x^3/2 wrt x is , 2/5x^5/2 . Therefore ,

\longrightarrow I = 5 \bigg( \dfrac{2}{5} x^{\frac{5}{2}} \bigg] ^1_0 \bigg)

On simplifying we will get ,

\longrightarrow \underline{\underline{ I = 2 }}

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The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow
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Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

H_0: \mu = 13252

At the alternative hypothesis, we test if the mean cost has increased, that is:

H_1: \mu > 13252

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that \mu = 13252

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This means that n = 20, X = 15560, s = 3500

Value of the test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

t = 2.95

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

7 0
3 years ago
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Answer:

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Step-by-step explanation:

4x=-5

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3 years ago
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Answer:

<em>Answer: B San Francisco</em>

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Check the picture below.

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3 years ago
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