Anybody needs help with their work, any subject

Suppose you regress the natural log of total family expenditures on household size using observations of households from the 201

postnew [5]

Answer:

Step-by-step explanation:

Please check below in the attached, you will see answer to given questions, thank you, I hope it helps.

4 0

3 years ago

Teresa earned scores of 88, 84, 88 points and 91 points on 4 test what is the lowest score she can get on the fifth test and sti

Usimov [2.4K]

Answer:

99 points

Step-by-step explanation:

Average of a data a is the sum of the data divided by the number of data.

Average = sum of data ÷ number of data

A = T/n .....1

For the average of 5 test to be 90.

The total sum of score must be;

Number of tests n = 5

Average A = 90

Substituting into equation 1

90 = T/5

T = 90×5 = 450 points

The total score must be equal to 450 point to have an average of 90 points

The sum of the first four test is;

88+84+88+91 = 351

Let x represent the first score,

351 + x = 450

x = 450 - 351

x = 99 points

The minimum score she can get in the fifth test is 99 points

5 0

3 years ago

During a concert the temperature should not rise above 40°C.

zaharov [31]

Answer:

the answer is 55

Step-by-step explanation:

if you subtract 40 degrees from 95 degrees you get 55 degrees. There are plenty of other ways to do that problem but i have a little bit of time. SORRY!

- I hope that helped

6 0

3 years ago

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i

USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

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:

$n = 55, \pi = \frac{24}{55} = 0.4364$

93% confidence level

So $\alpha = 0.07$ , z is the value of Z that has a pvalue of $1 - \frac{0.07}{2} = 0.965$ , so $Z = 1.81$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574$

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0

3 years ago

Questlon 10 of 10

Juli2301 [7.4K]

The answer is 75 degrees. These are supplementary angles so
180-105 = 75

8 0

3 years ago