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3 years ago

If , what is the truncation error for S4?

Mathematics

2 answers:

serious [3.7K]3 years ago

7 0

It is 0.037 so that mean you are right but no right

allsm [11]3 years ago

6 0

Answer:

A. .035

Step-by-step explanation:

Right on Edge

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What are the x-intercepts of this function? g(x) = -0.25x²-0.25x + 5

Nat2105 [25]

The x intercepts of the equation are -5 and 4.

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2 years ago

Factor -2.9 out of 8.7b – 11.6. Using gfc or without

Andreas93 [3]

Answer:

Im losted

Step-by-step explanation:

6 0

3 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

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

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, 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 at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, 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 significance level of 0.05 and 18 - 1 = 17 df.

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

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, 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 do not reject the null hypothesis, 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

3 0

3 years ago

A certain substance in an experiment was being stored at −3.2°F. It was then placed on a table where the temperature was raised

tatuchka [14]

Answer:

−0.6°F

Step-by-step explanation:

3 0

3 years ago

An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 8 sees

zheka24 [161]

Answer:

Probability = 0.12025

Step-by-step explanation:

P (Am) = 1/50 = 0.02 {Magazine ad}

P (At) = 1/8 = 0.125 {Television ad}

P (Am ∩ At ) = 1/100 = 0.01 {Both ads}

P (Am U At) = P (Am) + P (At) - P (Am ∩ At )

= 0.02 + 0.125 - 0.01

P (Am U At) = 0.135 {Person sees either ad}

P (Am' ∩ At') = 1 - P (Am U At)

P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}

Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)

P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1

P (P) = (0.25) (0.135) + (0.1) (0.865)

= 0.03375 + 0.0865

0.12025

7 0

3 years ago