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jok3333 [9.3K]
2 years ago
6

Alex bought 7.25 pounds of produce. If the produce was priced at $2.12 per pound, how much did Alex pay for the produce? How muc

h change did he receive if he paid with $20?
Mathematics
2 answers:
Leviafan [203]2 years ago
6 0

Answer:

Alex would need to pay $15.37 for the produce. If Alex payed with $20 he would receive $4.63 as change.

Step-by-step explanation:

Agata [3.3K]2 years ago
3 0
He paid $15.37 for the produce and would get $4.63 in return if he paid with $20
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Grayson is a kickboxing instructor who will be teaching classes at a local gym. To get certified as an instructor, he spent a to
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Answer:

If Jeffrey takes 50 classes in his first month as an instructor, he will earn back the amount he spent on certification.

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Suppose that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage
Zigmanuir [339]

Answer:

We conclude that the percentage of blue candies is equal to 29​%.

Step-by-step explanation:

We are given that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage of blue candies is equal to 29​%.

Let p = <u><em>population percentage of blue candies</em></u>

So, Null Hypothesis, H_0 : p = 29%     {means that the percentage of blue candies is equal to 29​%}

Alternate Hypothesis, H_A : p \neq 29%     {means that the percentage of blue candies is different from 29​%}

The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;

                         T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of blue coloured candies = 28%

           n = sample of colored​ candies = 100

So, <u><em>the test statistics</em></u> =  \frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }

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<u>Also, the P-value of the test statistics is given by;</u>

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Since the value of our test statistics lies within the range of critical values of z, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29​%.

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3 years ago
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Hello there!


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