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saul85 [17]
3 years ago
9

A retailer returned 2700 defective items to the manufacturer and received a credit for the retail price of $0.83 or 834/34 per i

tem less a discount of of the retail price. What was the amount of the credit received by the retailer?
Mathematics
1 answer:
Helga [31]3 years ago
8 0

Answer:

credit gain by retailer is $2257.39

Step-by-step explanation:

Given data:

Retaile price 83 4/3 c or 84.33c

Let Discount rate 6/7

Discount amount = (6/7)% of 84.33 c = 0.7228 C

Net credit received = 84.33c - 0.7228c = 83.607c

Net credit received = 83.607 c

Item required  = 2700

Total discount = 2700 × 83.0607 = 225739.44c

Total discount = $2257.39

credit received by retailer is $2257.39

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According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou
Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

p=X/n=35/140=0.25

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239

We calculate the z-score as:

z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

p=X/n=21/140=0.15

We can calculate then the z-score as:

z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

P(X

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231

We can calculate the z-scores for p1=0.18 and p2=0.25:

z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

P=P(0.18

5 0
3 years ago
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