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saul85 [17]
1 year ago
9

Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of s

amples in her neighborhood. It takes one person \dfrac{2}{3} 3 2 ​ start fraction, 2, divided by, 3, end fraction of a minute to prepare one bag. How many hours will it take Ashley and 444 of her friends to prepare 157515751575 bags of samples?
Mathematics
2 answers:
nikitadnepr [17]1 year ago
7 0

Answer:

3 1/2

Step-by-step explanation:

jasenka [17]1 year ago
6 0

Answer:

  3 1/2 hours

Step-by-step explanation:

This is a problem in units conversion. We want to get from bags to hours by way of minutes per bag. One bag takes an effort of 2/3 person·minute, so we need to divide the total effort by the number of persons and convert minutes to hours.

  (1575 bags)×(2/3 person·min/bag)/(5 person)/(60 min/h)

  = (1575)(2/3)(1/5)(1/60) h = 3.5

It will take the 5 of them about 3 1/2 hours to prepare 1575 bags.

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Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
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Answer:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

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A sports club needs to raise at least $500 by selling chocolate bars for $2.50 each. Sebastian wants to know how many chocolate
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Answer:

The inequality should be written as $2.50 times c is greater than or equal to c.

Step-by-step explanation:

First, identify what you know:

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2) Sebastian needs to raise at least $500

So, Sebastian needs to sell at least enough chocolate bars to hit $500. The inequality cannot be written as less than or equal to, because he can't sell less than the number of chocolate bars needed to make $500.

Automatically, I can calculate the minimum number of bars he'll need to sell.

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c must equal greater than or equal to $500, for Sebastian to raise enough money! So, basically Sebastian has to sell 200 bars OR more.

Hope this helps! :)

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