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S_A_V [24]
3 years ago
9

Suppose that the functions r and s are defined for all real numbers x as follows. =rx−x5 =sx+2x3 write the expressions for ·rsx

Mathematics
1 answer:
PolarNik [594]3 years ago
8 0

Although the question's print is not clear, to the best of my knowledge, as I can see, We are given that:

r(x)=-x^5 and s(x)=2x^3

No, we know that, by definition:

(r.s)(x)=r(x)\times s(x)

\therefore (r.s)(x)=-x^5\times 2x^3=-2x^8

Thus, for the functions r and s which are defined for all real numbers x, the answer to the question asked is:

(r.s)(x)=-2x^8

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Testing the hypothesis, it is found that:

a)

The null hypothesis is: H_0: \mu \leq 10

The alternative hypothesis is: H_1: \mu > 10

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The critical value is: t_c = 1.74

The decision rule is:

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Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

H_0: \mu \leq 10

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

H_1: \mu > 10

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: t_c = 1.74.

Hence, the decision rule is:

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Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem, we have that:

\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18

Thus, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}

t = 1.41

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

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