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2 years ago

Question 7 of 10 What is the slope of the line shown below? A. 2 B.-2 C.4 D.-4

Mathematics

2 answers:

Fofino [41]2 years ago

8 0

C.4 you need to use the formula for slope

Ira Lisetskai [31]2 years ago

4 0

Answer:

C. 4

Step-by-step explanation:

Use the slope formula y2-y1 / x2-x1

6+2/3-1

8/2

So, the answer is C. 4.

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Pedro thinks that he has a special relationship with the number 3. In particular, Pedro thinks that he would roll a 3 with a fai

anyanavicka [17]

Answer:

p value = 0.3122

Step-by-step explanation:

Given that Pedro thinks that he has a special relationship with the number 3. In particular,

Normally for a die to show 3, probability p = $\frac{1}{6} =0.1667$

Or proportion p = 0.1667

Pedro claims that this probability is more than 0.1667

$H_0 = P =0.1667______H_a = P>0.1667_____$

where P is the sample proportion.

b) n=30 and $P = 6/30 =0.20$

Mean difference = $0.2-0.1667=0.0333$

Std error for proportion = $\sqrt{\frac{p(1-p)}{n} } \\=0.0681$

Test statistic Z = p difference/std error = 0.4894

p value = 0.3122

p >0.05

So not significant difference between the two proportions.

7 0

2 years ago

While walking past a flag pole, Tommy looked down and noticed his shadow. Use the diagram below to

SashulF [63]

Answer: x=5.6

Step-by-step explanation:

3 0

2 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

2 years ago

Ana can finish a certain job in 10 hours. Together, Ana and Nina can finish the same job in 3 hours. How many hours can Nina fin

Over [174]

Answer: 4.5 hours

Step-by-step explanation:

$Ana: \dfrac{1}{10}\ \text{job per hour}\\\\Nina: \dfrac{1}{x}\ \text{job per hour}\\\\Together: \dfrac{1}{3}\ \text{job per hour}\\\\\\Ana + Nina = Together\\\dfrac{1}{10}\quad +\dfrac{1}{x}\quad =\quad \dfrac{1}{3}\\\\\\\dfrac{1}{10}(30x)+\dfrac{1}{x}(30x) =\dfrac{1}{3}(30x)\\\\\\\rightarrow\quad 3x+30=10x\\\\.\quad 3x+30=10x\\\underline{-3x\qquad \quad -3x\quad}\\.\qquad \quad 30=7x\\\\.\qquad \quad \dfrac{30}{7}=\dfrac{7x}{7}\\\\\\.\qquad \quad \dfrac{30}{7}=x$