PLEASE HELP- (-2, -7) (7, 2) (2, 7) (7, -2)

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago

3 connecting lines are shown. Line D F is horizontal. Line D E is about half the length of line D F. Line F E is about one-third

mars1129 [50]

The inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

<h3>Inequalities </h3>

From the question, we are to determine which of the given inequalities explains why the three segments cannot be used to construct a triangle

From the given information,

Line DE is about half the length of line DF

That is,

ED = 1/2 DF

Also,

Line FE is about one-third of the length of line DF

That is,

EF = 1/3 DF

Then, we can write that

ED + EF = 1/2DF + 1/3DF

ED + EF = 5/6 DF

Since,

5/6 DF < DF

Then,

ED + EF < DF

Hence, the inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

Learn more on Inequalities here: brainly.com/question/1447311

#SPJ1

5 0

2 years ago

6-y-x^2 use x=5 and y=4

Ulleksa [173]

Answer:

6 - 4 - 5 ^ 2 = -23

hope this helps

5 0

3 years ago

In a survey of 1000 large corporations, 260 said that, given a choice between a job candidate who smokes and an equally qualifie

Usimov [2.4K]

Answer:

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.