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Sonja [21]
3 years ago
12

A game involves rolling a fair six-sided die. If the number obtained on the die is a multiple of three, the player wins an amoun

t equal to the number on the die times $20. If the number is not a multiple of three, the player gets nothing.
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

Winning probability is \frac{1}{3}

Step-by-step explanation:

Favourable outcomes to win are 3 and 6.

Total outcomes are six in fair dice.

So , probability of winning some money =\frac{2}{6}=\frac{1}{3}

To win $20*3 or $20*6 favourable outcome is 1 (3 and 6 respectively)

Thus probability =\frac{1}{6}

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What is the value of 43 = 23? 2 7 8 16​
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3 0
2 years ago
Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

5 0
2 years ago
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