The Arnold family arrived at the beach at

ANSWER THIS QUESTION ASAP DONT SEND A FILE AT ALL. WHATS THE ANSWER I HAVE A TIKE LIMIT HELPP ONLY ANSWER IF YOU KNOW THE ANSWER

vfiekz [6]

Answer:reflect across

Step-by-step explanation:

4 0

3 years ago

2 pts

katrin2010 [14]

Answer:

It is a binomial and the degree is 9

4 0

3 years ago

What is the best estimate for the sum 2/9 + 10/11 ?

Gelneren [198K]

Answer:

4. 0 + 1 = 1

Step-by-step explanation:

since 2/9 is a bit small

and

since 10/11 is almost whole

3 0

3 years ago

Read 2 more answers

When Rachel put gas in her car she records the distance and she has driven since her last fill up and then number of gallon of g

katen-ka-za [31]

Answer:

You obviously copied the question text in an incomplete and lazy way.

I still really wanna help you on your problem.

Please either point the whole question with the possible answers or make a photograph of the problem.

7 0

3 years ago

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

4 years ago