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marin [14]
3 years ago
15

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be ch

osen if two people who are married to each other cannot both serve on the committee?
A. 16
B. 24
C. 26
D. 30
E. 32
Mathematics
1 answer:
coldgirl [10]3 years ago
8 0

total selections = 56

let's say that couple is always present in this committee of three.

This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways)

The third person can be selected out of remaining 6 people in 6 ways.

Therefore when couple exists there are: 4X6 = 24 ways

Thus no couple = (4X6) = 32

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Your answer is A

It cannot be C because the line y > -2  uses the (>) greater than symbol.

When equations have a greater than symbol, they are graphed using a <u>dotted line. </u>

It cannot be D because the line is  y ≤ - l x-1 l

with the less than or equal to sign (≤) , you use a <u>solid line</u> to graph


With the<u> greater than</u> symbol for y > -2 , the shaded region must be <em>above </em>this line.

With the <em>less than </em>or equal to sign in y ≤ - l x-1 l , the shaded region must be <em>below </em>the line.


Hope I helped - message me if you have any questions :)

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Answer:

x = -5/2 + i√19 and x = -5/2 - i√19

Step-by-step explanation:

Next time, please share the possible answer choices.

Here we can actually find the roots, using the quadratic formula or some other approach.

a = 1, b = 5 and c = 11.  Then the discriminant is b^2-4ac, or 5^2-4(1)(11).  Since the discriminant is negative, the roots are complex.  The discriminant value is 25-44, or -19.

Thus, the roots of the given poly are:

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x = -----------------------------------

                     2(1)

or x = -5/2 + i√19 and x = -5/2 - i√19

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3 years ago
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