7 9/16÷2 3/4<br><br><br> A. 214<br><br> B. 234<br><br> C. 414<br><br> D. 41316

1. Draw an appropriate Venn diagram to depict each of the following sets. U = The set of integers. A = The set of even integers.

CaHeK987 [17]

The attached diagram represents the Venn diagram of the sets

<h3>How to draw the Venn diagram?</h3>

The sets are given as:

The universal set, U = The set of integers.
A = The set of even integers.
B = The set of odd integers.
C = The set of multiples of 3.
D= The set of prime numbers

From the above representation, we have the following highlights:

Set A and set B will not intersect, because no number can be even and odd
Set C and set D will intersect set A because they have common elements 6 and 2, respectively
Set C and set D will intersect set B because they have common elements 3 and 3, respectively

Using the above highlights, we can now draw the Venn diagram

See attachment for the Venn diagram

Read more about Venn diagram at:

brainly.com/question/4910584

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3 0

2 years ago

What is another way to write -3 x (4 + 7)

KIM [24]

Negative 3 and the product of 4 more than 7

4 0

3 years ago

As a member of a music club, you can order CDs for $14.99 each. The music club also charges $4.99 for each shipment. The express

Dafna1 [17]

Howdy!

using the expression 14.99(n) + 4.99, the total of 3 CDs would cost $49.96.

8 0

3 years ago

Read 2 more answers

A baseball team practices 4 times a week. Each practice lasts for 115 minutes. The team practiced for 12 weeks during the season

Xelga [282]

Answer:

5520 min

Step-by-step explanation:

4*115=460

460*12=5520

4 0

3 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 $\pi = 0.5$