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Natali5045456 [20]
2 years ago
14

A game is played as follows: a die numbered 1 to 6 is rolled. If the number 1 comes up, you get nothing. If the number 2, 3, or

5 comes up, you win $4. If the number 4 or 6 comes up, you lose $6. According to the expected value, this is a fair game. True False
Mathematics
1 answer:
Komok [63]2 years ago
7 0
<h2>Steps</h2>

So before I get to the steps to solving, we should first define what a fair game is. A fair game is a game where there is equal chance of win or loss, so <u>for this to be a fair game, the expected value must be zero.</u>

Now that we defined what a fair game is, let's get to solving! Firstly, we want to get all the possibilities and multiply their outcomes with their chance. In this case the possibilities are {1,2,3,4,5,6}, all have a 1/6 chance, and {1} outcome is 0, {2,3,5} outcome is 4 and {4,6} outcome is -6:

\textsf{1:}\ 0\times \frac{1}{6}=0\\\\\textsf{2:}\ 4\times \frac{1}{6}=\frac{4}{6}\\\\\textsf{3:}\ 4\times \frac{1}{6}=\frac{4}{6}\\\\\textsf{4:}\ -6\times \frac{1}{6}=-\frac{6}{6}\\\\\textsf{5:}\ 4\times \frac{1}{6}=\frac{4}{6}\\\\\textsf{6:}\ -6\times \frac{1}{6}=-\frac{6}{6}

Next, take all the products and add them up:

\frac{0}{6}+\frac{4}{6}+\frac{4}{6}-\frac{6}{6}+\frac{4}{6}-\frac{6}{6}=\frac{0}{6}=0

<h2>Answer</h2>

Since the expected value is zero, <u>it is true that this is a fair game.</u>

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A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

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