7. Craig rents cars. He earns $75.00 per day plus

No body helped me the last time so I’m posting it again I really need help with this

PIT_PIT [208]

B is equal to 154 degrees. we know this cause half of the circle equals 180 and if you subtract a (26) from 180 you get 154

3 0

3 years ago

I need help really bad on 1 and 2

Dmitry [639]

For question one:
First class goes from 8:00 to 9:00 second class starts 10 mins later at 9:10 and goes to 10:10. Ten minute break again 3rd class starts at 10:20, goes to 11:20. 10 min break again, fourth class starts 10 mins later at 11:30 and goes to 12:30. The answer for 1 is 12:30

8 0

3 years ago

Your credit card has a balance of $5400 and an annual interest rate of 17%. You decide to pay off the balance over two years. If

Phoenix [80]

Answer:

5400/12= 450 paid

1006.56/2=503.28 interest

total is 5,903.28

total for paid 491.94 for months

Step-by-step explanation:

6 0

3 years ago

5. An airplane takes 4 hr to travel a distance of 4100 km. Another airplane travels at a speed

Mazyrski [523]

Answer:

4.969696 hours

Step-by-step explanation:

The distance traveled is shown through the following equations

Distance = velocity * time

This means that

4100=velocity*4

This means that ariplane 1 travles at 1025 km/hr

We can then subtract 200 from this to find that airplane two travels at a speed of 825 km/hr

Now we need to find how long it takes for that plane to travel 4100

So using the same equation

4100=825*time

Divison will tell us that the answer is around 4.9696 hours which is pretty much 5

7 0

3 years ago

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 $\pi = 0.5$