PLS HELP 10 POINTS ASAP RN PLSSSSS OR I WILL FAIL CLASS

Identify the domain of the function

sertanlavr [38]

The answer is C. is right

7 0

3 years ago

Read 2 more answers

Simplify\:\frac{\sqrt{5}}{\sqrt{3}}

avanturin [10]

$\dfrac{\sqrt 5}{\sqrt 3} =\sqrt{\dfrac 53} \approx 1.291$

7 0

3 years ago

Line t has an equation of y = -8x + 7. Line u includes the point (-1, 7) and is parallel to line

umka2103 [35]

Answer:

y = -8x - 1

Step-by-step explanation:

y = mx + b

7 = -8 (-1) + b Substitute in points for x and y. Substitute -8 for m (parallel)

7 = 8 + b

-1 = b

8 0

3 years ago

A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true

kodGreya [7K]

Answer:

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 542, \pi = \frac{301}{542} = 0.555$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 - 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.52$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 + 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.59$

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

5 0

3 years ago

Chi square Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, perc

Marat540 [252]

Answer:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

Step-by-step explanation:

The missing data for the random sample of 80 days between December and April with snowfall is:

Month Days

December 16

January 11

February 16

March 18

April 19

The Chi-square goodness of fit test would be used to determine whether the snow falls in Daphne's hometown followed the given distribution.

The hypothesis for the test can be defined as follows:

H₀: The snow falls in Daphne's hometown does not follow the given distribution.

Hₐ: The snow falls in Daphne's hometown followed the given distribution.

Assume that the significance level of the test is, α = 0.05.

The Chi-square test statistic is given by:

$\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}$

The value of Chi-square test statistic is computed in the Excel sheet below.

$\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}=8.383$

Compute the p-value as follows:

$\text{p-value}=P(\chi^{2}_{(4)}>8.383)=CHISQ.DIST.RT(8.383,4)=0.0785$

The p-value of the test is 0.0785.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

p-value = 0.0785 > α = 0.05.

The null hypothesis will not be rejected.

Conclusion:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

4 0

4 years ago