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wolverine [178]
3 years ago
7

Coffee beans worth $ 10.50 $10.50 per pound are to be mixed with coffee beans worth $ 13.00 $13.00 per pound to make 50 50 pound

s of beans worth $ 11.50 $11.50 per pound. How many pounds of each type of coffee bean should be used?
Mathematics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer: 30 pounds of coffee beans worth $10.50 per pound was used.

20 pounds of coffee beans worth $13.00 per pound was used.

Step-by-step explanation:

Let x represent the number of coffee beans worth $10.50 per pound that should be mixed.

Let y represent the number of coffee beans worth $13.00 per pound that should be mixed.

50 pounds of beans worth $11.50 per pound is to be made. The total cost of the mixture would be

50 × 11.5 = 575

This means that

10.5x + 13y = 575 - - - - - - - 1

The total number of pounds of each type of coffee used is 50. This means that

x + y = 50

Substituting x = 50 - y into equation 1, it becomes.

10.5(50 - y) + 13y = 575

525 - 10.5y + 13y = 575

- 10.5y + 13y = 575 - 525

2.5y = 50

y = 50/2.5 = 20

x = 50 - 20= 30

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Step-by-step explanation:

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Suppose you are asked to take a sample of men and women in order to measure their emotional IQ. The emotional IQ scale that you
yan [13]

Answer:

The 99% confidence interval for the scores of men is (64.406,75.594).

The 99% confidence interval for the scores of women is (73.609,90.391).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution will be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.797.

For men:

Standard deviation of 10.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.797\frac{10}{\sqrt{25}} = 5.594

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 70 - 5.594 = 64.406

The upper end of the interval is the sample mean added to M. So it is 70 + 5.594 = 75.594

The 99% confidence interval for the scores of men is (64.406,75.594).

For women:

Standard deviation of 15.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.797\frac{15}{\sqrt{25}} = 8.391

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 82 - 8.391 = 73.609

The upper end of the interval is the sample mean added to M. So it is 82 + 8.391 = 90.391

The 99% confidence interval for the scores of women is (73.609,90.391).

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Step-by-step explanation:

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Answer:

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Answer:

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