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madam [21]
3 years ago
11

An arts supply store has 28 pounds of modeling clay to ship to a client. Each packing box can contain no more than

Mathematics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

D. The store will need exactly 38 boxes for the shipment, with  1/3 pound of clay going into the last box

Step-by-step explanation:

We need to divide the number of pounds by the numbers of pounds per box

28 ÷ 3/4

Copy dot flip

28 * 4/3

112/3

37 1/3

We will have to round up to pack all the clay

38 boxes with 1/3 in the 38th box

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Pls help me on this!
Zigmanuir [339]

Answer:

4 45/100 Keep simplifying it. You can do it!

Step-by-step explanation:

4 is the whole. 45 is out of 100

7 0
3 years ago
investments historically have doubled every 9 years. If you start will a $2,000 investment, how much money would you have after
azamat

Answer:

$64,000

Step-by-step explanation:

Friday, 45 / 9 = 5

So the investment will double 5 times

First time

2,000 (starting investment) × 2 = 4000

Second time

4000 × 2 = 8000

Third time

8000 × 2 = 16000

Fourth time

16000 × 2 = 32000

Fifth and final time

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$64,000

7 0
2 years ago
3/8 divided by 1/2 equals what
yulyashka [42]
0.75. This is the correct answer
7 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

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

0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.009\sqrt{n} = 2.575*0.5

\sqrt{n} = \frac{2.575*0.5}{0.009}

(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2

n = 20464.9

Rounding up:

A sample of 20465 is required.

8 0
2 years ago
If you know how to do this, please write an answer.
daser333 [38]

hint the circle is 40 inches and the square is 100 inches

5 0
3 years ago
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