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Dovator [93]
3 years ago
6

If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each o

ther?
Mathematics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

16/33

Step-by-step explanation:

12C4(total possible outcomes)=495

Since there are 6 married couples there must be 6*[2C2×10C2]=270

must subtracted 15 since when there are 2 couples in a group would have doubled 270-15=255

the no of groupings with no married couples 495-255 =240

probability=no of favourable outcomes /no of possible outcome

probability=240/495= 16/33

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Mathematical
Ivenika [448]

Answer:

7

Step-by-step explanation:

32 - 2 = 30

30 - 2 = 28

From those 28 left that Sarah has left, divide by 4 to get 7 in each group remaining.

4 0
2 years ago
Read 2 more answers
Evaluate the expression for the given value of the variable: 7(m-6) when m = 8
tatuchka [14]

Answer:

14

Step-by-step explanation:

7( 8 - 6 ) = 7 ( 2 ) = 14

Following PEMDAS, you would begin with solving within the ( ). So being that m = 8 you plug that in. Then subtract 6 from 8 which would then give you 2. The multiply 7 by 2, which gives you 14.

4 0
3 years ago
Use the distributive property to remove the parentheses. -6(11 - 2)​
salantis [7]

Answer:

-66+12 or -54

Step-by-step explanation:

To remove the parenthesis multiply the number outside of the parenthesis (in this case -6) to the numbers inside of the parenthesis (11, and -2).

-6*11=-66

A negative multiplied by a positive is a negative.

-6*-2=12

A negative multiplied by a negative is a positive.

-66+12=-54

7 0
3 years ago
Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5
Marianna [84]
\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}

\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.
7 0
3 years ago
An inequality is a statement that compares two expressions that are strictly equal.
djyliett [7]

Answer:

False

Step-by-step explanation:

an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions.

5 0
3 years ago
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