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natima [27]
2 years ago
10

Which is a better buy? 3-pound bag of candy for $6.04 or 9-pound bag of candy for $18.45?

Mathematics
2 answers:
andrezito [222]2 years ago
7 0
The first option is the correct answer

Explanation
Multiply the 3 pound bag value by 3 giving you 18.12 which is less than the 9 pound bag
mylen [45]2 years ago
7 0

Answer:

the 3 pound bag

Step-by-step explanation:

6.04÷3= $2.013 per pond

and

18.45÷9= $2.05 per pound

You want to pick the lesser value because that means it will cost you less.

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7987.1569 to the nearest thousandth
natima [27]

Answer:

7987.1569 to the nearest thousandths is 7987.157

Step-by-step explanation:

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Simplify the expression 9w+w+5w
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"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

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