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ella [17]
3 years ago
13

It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of repondent

s reported that they had participated in an online auction. In this same survey, 45% of repondents were men and 38% were men who had participated in online auctions. What is the probability that a respondent selected at random is female and has never participated in an online auction?
A. 0.62
B. 0.72
C. 0.38
D. 0.28
E. 0.07
Mathematics
1 answer:
kupik [55]3 years ago
5 0

Answer:

Hence, the Female and not participated is \frac{28}{100}=0.28\%

Step-by-step explanation:

Let Total respondents =100

Participated in online action (P)=65

Men (m)=45  and Women (F)=100-45=55

Male and participated (M\; \text{and}\;P)=38

Female and participated (F\; \text{and}\;P)=65-38=27

Male and not participated is 45-38=7

Female and not participated is 55-27=28

Therefore, the Female and not participated is \frac{28}{100}=0.28\%

Hence, the correct option is D.

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Yuri purchased 8 trees to have planted at his house. The store charged a delivery fee of $5 per tree.
cestrela7 [59]

Answer:

$16.

Step-by-step explanation:

Let t represent the cost of each tree.

We have been given that Yuri purchased 8 trees to have planted at his house. So the cost of 8 trees will be 8t.

We are also told that the store charged a delivery fee of $5 per tree. So the delivery charges of 8 trees will be 8\times \$5=\$40

The total cost of purchasing 8 trees will be 8t+40.  

Since the total cost of the trees including the delivery was $168, so we can get an equation as:

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Therefore, the cost of each tree is $16.

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3 years ago
A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The
TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 16 ​salmon, and to discount outlet stores in pallets of 64 salmon. To forecast​ costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts​ (a) and​ (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: \frac{s.d}{\sqrt{u}}

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{9}}

= \frac{3}{3} = 1

The standard deviation = 1

ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{16}}

= \frac{3}{4} = 0.75

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{64}}

= \frac{3}{8} = 0.375

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

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Answer:

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Step-by-step explanation:

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90 + 15 = 105

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