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ValentinkaMS [17]
3 years ago
14

Katherine went to the store to buy some almonds. The price per pound of the almonds is $3 per pound and she has a coupon for $1.

50 off the final amount. With the coupon, how much would Katherine have to pay to buy 4 pounds of almonds? Also, write an expression for the cost to buy pp pounds of almonds, assuming at least one pound is purchased.
Cost of 4 pounds:
Cost of pp pounds:
Mathematics
2 answers:
salantis [7]3 years ago
7 0

Answer: 12

Step-by-step explanation:

4t5t

Skylar went to the store to buy some almonds. The price per pound of the almonds is $7.75 per pound and she has a coupon for $3.50 off the final amount. With the coupon, how much would Skylar have to pay to buy 3 pounds of almonds? Also, write an expression for the cost to buy pp pounds of almonds, assuming at least one pound is purchased.

Skylar went to the store to buy some almonds. The price per pound of the almonds is $7.75 per pound and she has a coupon for $3.50 off the final amount. With the coupon, how much would Skylar have to pay to buy 3 pounds of almonds? Also, write an expression for the cost to buy pp pounds of almonds, assuming at least one pound is purchased.

Skylar went to the store to buy some almonds. The price per pound of the almonds is $7.75 per pound and she has a coupon for $3.50 off the final amount. With the coupon, how much would Skylar have to pay to buy 3 pounds of almonds? Also, write an expression for the cost to buy pp pounds of almonds, assuming at least one pound is purchased.

kolbaska11 [484]3 years ago
6 0

Answer:10.50

Step-by-step explanation: 12-1.50=10.50

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6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul
drek231 [11]

Answer:

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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 zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is of:

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

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that n = 1012, \pi = \frac{374}{1012} = 0.3696

95% confidence level

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

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}

E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}

\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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