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Elan Coil [88]
3 years ago
11

I can’t figure out this geometry question BFG=_____

Mathematics
1 answer:
lawyer [7]3 years ago
6 0
M<BFG = 1/2(arcGB + arcCD)

answer is A. first one.
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3 years ago
Consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number,
Aleksandr-060686 [28]

Answer:

The expected value of the points earned on a single roll in this game is \dfrac{1}{6} = 0.1667 .

Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

Note that the probability of getting any of the specific number on the dice is   \dfrac{1}{6} .

      Numbers on the dice (X)                       P(X)

                      +1                                                 \frac{1}{6}

                      +2                                                \frac{1}{6}

                      +3                                                \frac{1}{6}

                      -4                                                 \frac{1}{6}

                      +5                                                \frac{1}{6}

                      -6                                                 \frac{1}{6}

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X)  =  \sum X \times P(X)

   =  (+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}

   =  \frac{1}{6} + \frac{2}{6} + \frac{3}{6} - \frac{4}{6} + \frac{5}{6} - \frac{6}{6}

   =  \frac{1+2+3-4+5-6}{6}

   =  \frac{11-10}{6}= \frac{1}{6}

Hence, the expected value of the points earned on a single roll in this game is  \frac{1}{6} = 0.1667 .

4 0
3 years ago
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