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Cerrena [4.2K]
4 years ago
7

Suppose that each time Diana Taurasi takes a 3-point shot, she has a 37% probability of success, independent of all other attemp

ts. (Success yields 3 points; failure yields 0.) If Taurasi takes seven 3-point shots in a game, what is the variance of the total number of points she scores from these shots?
Mathematics
1 answer:
Ivanshal [37]4 years ago
8 0

Answer:

5 {}^{2}  + (8 - 5) \div 2 \:

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What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?
Ahat [919]

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\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}

So the quotient is 1x^2+5x+-1 and the remainder is -6.

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form x-c.

Since we are dividing by x+2=x-(-2), this is our setup for the synthetic division:

-2 |     1    7    9    -8

   |         -2   -10     2

   ______________

          1   5    -1       -6

So the quotient is 1x^2+5x+-1 and the remainder is -6.

So \frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}.

We can also do long division.

                 x^2+5x-1

               ____________________

        x+2| x^3+7x^2+9x-8

              -(x^3+2x^2)

              -------------------

                       5x^2+9x-8

                    -( 5x^2+10x)

                      --------------------

                                 -x-8

                               -(-x-2)

                                --------------

                                      -6

So we see here we get the same quotient, x^2+5x-1. and the same remainder, -6.

Now let's check our result that:

\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}.

So I'm going to rewrite the right hand side as a single fraction:

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}.

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}

Now let's focus on multiplying (x+2)(x^2+5x-1).

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

x(x^2)=x^3

x(5x)=5x^2

x(-1)=-x

2(x^2)=2x^2

2(5x)=10x

2(-1)=-2

---------------------------Combine like terms:

x^3+(5x^2+2x^2)+(-x+10x)+-2

x^3+7x^2+9x-2

So let's go back where we were in our check of \frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}:

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}

\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}

\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}

We have the exact same thing on both sides so we did good.

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