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White raven [17]
2 years ago
13

(1 point) The players on a soccer team wear shirts, with each player having one of the numbers 1, 2, ..., 11 on their backs. The

set A contains players with even numbers on their shirts. The set B comprises players wearing an odd number less than 7. The set C contains the defenders, which are those wearing numbers less than 6. Select the correct set that corresponds to each of the following. Part a) A∩(B∪C) A. {1,2,3,4,5} B. ∅ C. {1,3,5} D. {2,4} E. {2} Part b) (A∩Bc)∪(B∩C)c A. {6,7,8,9,11} B. {2,4,6,7,8,9,10,11} C. {2,3,4,5,6,8,10} D. {1,2,3,4,5,6,8,10} E. {6,7,8,10,11}
Mathematics
1 answer:
AlekseyPX2 years ago
6 0

Answer:

Part a) D.

Part b) B.

Step-by-step explanation:

A = {2, 4, 6, 8, 10}

B = {1, 3, 5}

C = {1, 2, 3, 4, 5}

Part a)

A∩(B∪C) =

= {2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5}

= {2, 4}

Answer: D.

Part b)

(A∩Bc)∪(B∩C)c =

= ({2, 4, 6, 8, 10}∩{2, 4, 6, 7, 8, 9, 10, 11}) ∪ ({1, 3, 5}∩ {1, 2, 3, 4, 5})c

= {2, 4, 6, 8, 10} ∪ {1, 3, 5}c

= {2, 4, 6, 8, 10} ∪ {2, 4, 6, 7, 8, 9, 10, 11}

= {2, 4, 6, 7, 8, 9, 10, 11}

Answer: B

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Answer:

Step-by-step explanation:

5 tables for 35 bucks.....unit cost is 35/5 = $ 7 bucks per table

30 chairs for 60 bucks...unit cost is 60/30 = $ 2 bucks per chair

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Ratna , who is in Mumbai , telephones her brother Ravi in Ahmedabad and speaks for 89 seconds . She is charged Rs 1.20 for every
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16.40

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1 year ago
The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou
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Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = average annual amount American households spend on daily transportation = $6,312

           \sigma = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

                      P(X < $1,000) = 0.05

                      P( \frac{X-\mu}{\sigma} < \frac{\$1000-\$6312}{\sigma} ) = 0.05

                      P(Z < \frac{\$1000-\$6312}{\sigma} ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

                           \frac{\$1000-\$6312}{\sigma}=-1.645                

                            \sigma=\frac{-\$5312}{-1.645}  = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

      P($4000 < X < $6000) = P(X < $6000) - P(X \leq $4000)

 P(X < $6000) = P( \frac{X-\mu}{\sigma} < \frac{\$6000-\$6312}{\$3229.18} ) = P(Z < -0.09) = 1 - P(Z \leq 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X \leq $4000) = P( \frac{X-\mu}{\sigma} \leq \frac{\$4000-\$6312}{\$3229.18} ) = P(Z \leq -0.72) = 1 - P(Z < 0.72)

                                                            = 1 - 0.7642 = 0.2358  

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

                    P(X > x) = 0.03   {where x is the required range}

                    P( \frac{X-\mu}{\sigma} > \frac{x-\$6312}{3229.18} ) = 0.03

                    P(Z > \frac{x-\$6312}{3229.18} ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

                           \frac{x-\$6312}{3229.18}=1.88                

                         {x-\$6312}=1.88\times 3229.18  

                          x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

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Answer:

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We have 4 letters initially, so we can choose any 1 as our first letter. We have 4 choices for our first letter

However, once we choose our first letter, we can't use it anymore, so, for our second letter, we can only choose from the remaining 3 letters.

Furthermore, once we choose our second letter, we can only choose our 3rd letter from the remaining two letters we didn't choose yet.

Finally, our last letter will always be the one we didn't choose the last 3 times. So there is only one choice here.

Going off of this, we have four choices for the 1st letter, three choices for the 2nd letter, two choices for the 3rd letter, and one choice for the 4th letter

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In this case, it would be 4!

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