Leonel computed the average rate of change in the depth of a pool over a two week interval to be zero. Which statement is true

A cylindrical rain barrel has a radius of 2 feet and holds a total of 30 cubic feet of water. How tall is the rain barrel? Use 3

BabaBlast [244]

B. 2.39 feet. xoxo
pls rate my answer and mark me:))

3 0

3 years ago

Read 2 more answers

Please help! I am giving you 10 points for it!

uysha [10]

So if John gives a pen and pencil to each and all of them are over we know that the no. of pencils and pens should be the same
So we will find the LCM of 15 and 40 that is equal to 120
So John brought 120/15 that is 8 packets of pen and 120/40 that is three packets of pencils

6 0

3 years ago

Read 2 more answers

9 - 3 divided by 1/3 + 1

jolli1 [7]

Answer:

its just 9

Step-by-step explanation:

Divide: 3 : 1 / 3 = 3 / 1 · 3 / 1 = 3 · 3 / 1 · 1 = 9 / 1 = 9 Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator,

6 0

2 years ago

Read 2 more answers

An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

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 $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

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

A sample of n is required, and n is found for M = 0.009. So

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

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$