PLS HELP I WILL GIVE BRAINLIEST

Find all angle measures

Natali [406]

Answer:

1=90

2=68

3=112

4=112

5=68

6=68

7=112

8=112

3 0

3 years ago

david is 8 years older than stephen and 2 years younger than Graham. if Neil is 7 years younger than Graham, how much older than

ANEK [815]

5 years older...cause if David is 2 years younger than Graham & Neil is 7 years younger than Graham...7-2=5 years :)

6 0

3 years ago

Need help with this

Ganezh [65]

average = 85 MPH and traveled for 6 hours

total = 85 x 6 = 510 miles

answer

510 miles from Orlando to Augusta

4 0

4 years ago

Find the amount accumulated after investing a principle P for T years at an interest rate compounded k times per year.

BartSMP [9]

1. $3500(1+ \frac{.05}{4}) ^{4*10}$
$3500(1+.0125) ^{40}$
3500(1.64) = 5740

2. $25300(1+ \frac{.045}{12} ) ^{12*25}$
$25300(1+.00375) ^{300}$
25300(3.07) = 77671

Hope that helps. Feel free to ask any questions

8 0

4 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 $\pi = 0.5$