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GuDViN [60]
3 years ago
7

Suppose that, in an experimental setting, 100 students are asked to choose between Gamble A and Gamble B, where: Gamble A: The s

tudent will receive $5,100 with a 70 percent probability and $200 with a 30 percent probability. Gamble B: The student will receive $5,100 with a 50 percent probability, $200 with a 25 percent probability, and $0 (nothing) with a 25 percent probability. What is the expected value (EV) of Gamble B
Mathematics
1 answer:
Stolb23 [73]3 years ago
4 0
<h3>Answer: $2600</h3>

Focus on Gamble B only. Multiply each winnings with their corresponding probabilities.

5100*0.50 = 2550

200*0.25 = 50

0*0.25 = 0

Add up those results: 2550+50+0 = 2600

The expected value of gamble B is $2600

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pentagon [3]
For this case we have the following relationship:
 (x / 5280) = (7/100)
 From this relationship we must clear the value of x.
 Clearing we have:
 x = (7/100) * (5280)
 Calculating we have:
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it does rise in a horizontal distance of 1 mile about:
 
x = 369.6 feet
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3 years ago
Write a polynomial that represents the length of the rectangle<br><br> help please !!
disa [49]

Answer:

\textsf{Length}=0.8x^2-0.7x+0.7\quad \sf units

Step-by-step explanation:

Area of a rectangle = width × length

Therefore, to find the length of the rectangle, we need to divide the area by the width.

Using long division:

\large\begin{array}{r}0.8x^2-0.7x+0.7\phantom{)} \\x+0.5{\overline{\smash{\big)}\,0.8x^3-0.3x^2+0.35x+0.35\phantom{)}}}\\-~\phantom{(}\underline{(0.8x^3+0.4x^2)\phantom{-b))))))))))))))}}\\0-0.7x^2+0.35x+0.35\phantom{)}\\ \underline{-~\phantom{()}(-0.7x^2-0.35x)\phantom{-b)))))}}\\ 0.7x+0.35\phantom{)}\\\underline{-~\phantom{()}(0.7x-0.35)}\\ 0\phantom{)}\end{array}

Therefore, the <u>length</u> of the rectangle is:

0.8x^2-0.7x+0.7

5 0
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astraxan [27]
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xz_007 [3.2K]
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12 years

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20-8=12

Micheal's babysitter is 12 years older than Micheal.

8 0
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