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34kurt
1 year ago
7

Kayla is a basketball player who makes 50 of her 2 point shots and 20 of her 3 point shots find Kayla expected value for a 2 poi

nt shot
Mathematics
1 answer:
Mumz [18]1 year ago
4 0

Kayla's expected value for a 2- point shot is 1 point.

What is expected value?

The expected value, also known as expectation, expectancy, arithmetical expectation, mean, average, first or moment, is a generalization of the weighted average in the field of probability theory. Informally, the estimation is the arithmetic mean of a significant number of values of a random variable that were independently chosen. A weighted average of all potential outcomes constitutes the expected value of the a random variable with such a finite number of outcomes. When there is a continuum of potential outcomes, integration defines the expectation. The expectation is provided by Lebesgue integration in the measure theory-based axiomatic foundation for probability.

Sol- The expected value is calcuated using the formula:

EV= x.P(x)

where x is the weight/value of the event, and P(x) is the probability of the event occurring.

For a 2-point shot, we have:

X=2

P(x) = 50/100 = 0.5

There for the expected value will be -

EV = 2×0.5 = 1

Thus,

The expected value will be 1 point.

To know more about expected value click-
brainly.com/question/843074
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Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

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\begin{gathered} x-3=0 \\ x=3 \end{gathered}

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Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

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