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Vaselesa [24]
3 years ago
6

A farmer bought a number of pigs for $160. However, 8 of them died before he could sell the rest at a profit of 6 per pig. His t

otal profit was $8. How many pigs did he originally buy?
Mathematics
1 answer:
juin [17]3 years ago
6 0

Answer:

The farmer bought 36 pigs in all.

Step-by-step explanation:

Given that a farmer bought a number of pigs for $ 160, but, however, 8 of them died before he could sell the rest at a profit of 6 per pig, and his total profit was $ 8, to determine how many pigs did he originally buy the following calculation must be performed:

160 + 8 = 168

168/6 = 28

28 + 8 = 36

Thus, the farmer bought 36 pigs in all.

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Answer:

B) x+22

Step-by-step explanation:

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PS= x+22

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Factor.<br><br> 3x(y−4)−2(y−4)<br><br><br><br> Enter your answer in the box.
NISA [10]

Answer: Use the distributive property to multiply 3 by y−4.

3y−12−2(y−4)

Use the distributive property to multiply −2 by y−4.

3y−12−2y+8

Combine 3y and −2y to get y.

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Step-by-step explanation:

Hope this helps!

6 0
2 years ago
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon
fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that n = 603, \pi = \frac{142}{603} = 0.2355

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

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2 years ago
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Answer:

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Step-by-step explanation:

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