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Semenov [28]
3 years ago
8

at the movies, la quinta paid for drinks and popcorn for herself and her two children. she spend twice as much on popcorn as on

drinks. if her total bill came to $17.94, how much did she spend on drinks and how much did she spend on popcorn?
Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:

Step-by-step explanation:

let the amount spent on popcorn be x and amount spent on drinks be y

If the total bill paid is  $17.94, then x+y =  $17.94

If she spend twice as much on popcorn as on drinks, then y = 2x

Substitute y = 2x into the original equation

x+y =  $17.94

x + 2x =  $17.94

3x =  $17.94

x =  $17.94/3

x = $5.98

Since y = 2x

y = 2($5.98)

y = $11.96

Therefore she spent  $5.98 on popcorn and $11.96 on drinks

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b) The minimum sample size is 2401.

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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.

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The margin of error is:

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For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

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.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

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Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

The minimum sample size is 2401.

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