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Nookie1986 [14]
3 years ago
9

The academic computing committee at a college is in the process of evaluating different computer systems. The committee consists

of five ​administrators, seven ​faculty, and four students. A six person subcommittee is to be formed. The chair and vice chair of the committee must be​ administrators; the remainder of the committee will consist of faculty and students. In how many ways can this subcommittee be​ formed?
Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0
There are 3,300 different ways to select the committee.

First, we need to determine the total number of ways the 6 members can be selected.

For the chair and vice chair, it would be 5 x 4 = 20, but then that has to be divided by 2. (Because the order doesn't matter)
20 / 2 = 10

For the rest, it would be 11 x 10 x 9 x 8 = 7920, but again divide it by 4! because the order doesn't matter.
7920 / 24 = 330

To find the total ways for both, multiply 10 by 330 to get 3300.
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A certain brand of dinnerware set comes in three colors: red, white, and blue. Twenty percent of customers order the red set, 45
denis23 [38]

Answer:

a) 0.20

b) 0.45

c) 0.65

d) Yes

e) Yes

f) Z = X + Y (except when X = 1 and Y = 1)

This is because the successes of X and Y are mutually exclusive events but their failures aren't. X and Y cannot both be 1.

Step-by-step explanation:

Probability of a red set = 20% = 0.20

Probability of a white set = 45% = 0.45

Probability of a blue set = 35% = 0.35

Probability of the single set being a red or white set = 20% + 45% = 65% = 0.65

P(X=1) = 0.20, P(X=0) = 1 - 0.2 = 0.80

P(Y=1) = 0.45, P(Y=0) = 1 - 0.45 = 0.55

P(Z=1) = 0.65, P(Z=0) = 1 - 0.65 = 0.35

a) pX = P(X=1) = 0.20

b) pY = P(Y=1) = 0.45

c) pZ = P(Z=1) = 0.65

d) Since only one order is being considered at a time, it isn't possible to order red & white set in a single set, hence, both X and Y cannot both be successes (equal to 1) at the same time. But they can both be failures (both equal to 0) if a blue set is ordered. The successes of X and Y are mutually exclusive events but their failures aren't

e) Is pZ = pX + pY

pX = 0.2, pY = 0.45, pZ = 0.65

Hence, this statement is correct!

f) Z = X + Y

Let's check all the probabilities

when X = 1 and Y = 1, Z = 1

1 ≠ 1 + 1

when X = 0 and Y = 1, Z = 1

1 = 0 + 1

when X = 1 and Y = 0, Z = 1

1 = 1 + 0

when X = 0 and Y = 0, Z = 0

0 = 0 + 0

Hence, Z = X + Y (except when X = 1 and Y = 1)

This is because the success of X and Y are mutually exclusive events but their failures aren't.

8 0
3 years ago
Simplify the expression. Write your answer as a power.<br> <img src="https://tex.z-dn.net/?f=%285%5E%7B4%7D%20%29%5E%7B3%7D" id=
Charra [1.4K]

Answer:

5^{12}

Step-by-step explanation:

Lets say there's a number 'x' and there's an expression (x^a)^b where 'a' & 'b' are also numbers , <u>the simplified form of the expression is </u>x^{a \times b}<u>.</u>

So ,

(5^4)^3 = 5^{4 \times 3} = 5^{12}

6 0
2 years ago
Brad needs to rent scaffolding to paint a 2-story room. Handy Rentals charges $15 plus $12.50 per hour. The EZ Rental Company ch
mrs_skeptik [129]

Answer:

6 hours

Step-by-step explanation:

lmk if you want an explanation

6 0
2 years ago
The midpoint of AB is M(0,2). If the
otez555 [7]

Answer:

Step-by-step explanation:

(x+2)/2 = 0

x + 2 = 0

x = -2

(y - 3)/2 = 2

y - 3 = 4

y = 7

(-2, 7)

8 0
3 years ago
Line A is represented by the following equation: x + y = 2
jeka57 [31]
A system of linear equations will have no solution when the two lines making up the equation are parallel. Here, a system having x + y = 2 will have no solution if the second equation is x + y = a. where a is any real number.
5 0
3 years ago
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