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Amanda [17]
1 year ago
5

Sasha shared 20 muesli bars with three friends.

Mathematics
1 answer:
mestny [16]1 year ago
7 0

Answer: Sasha has 20 bars to start of with. Let's assume Sasha keeps 6 bars for herself. The problem says Sasha has twice as many bars as Hanna, so Hanna HAS to have 3 bars. The problem says Kyle got 3 more bars than Hanna, so 3+3=6, and Hannah got 2 less bars than Noah.. Noah has to have bars because 5-3=2

Sasha 6, Kyle 6, Noah 5, Hanna 3= 6+6+5+3=20

Step-by-step explanation:

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Use mental math. A resort has 380 rooms with space for 6 people. Estimate the greatest number of people who could be in these ro
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6 people

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Since the resort has 380 rooms with space for 6 people, only 6 people can be in these rooms at the same time.

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Irrespective of the number of rooms available, the space limit is 6, hence not more than 6 people can be accommodated.

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2 years ago
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16 inches by 36 inches

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Jordan wants to play a basketball game at a carnival. The game costs the player $5 dollar sign, 5 to play, and the player gets t
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The expected value of Jordan gains is -1 dollar.

Step-by-step explanation:

Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows

Y = 5 if X=2 (since he gets 10, but invested 5),  Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.

P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).

By definition, the expected value of Y is given by

E[Y] = 5\cdot P(Y=5)+0\cdot P(Y=0)-5 P(Y=-5). By the previous analysis, we have that

E[Y] = 5\cdot P(X=2)-5P(X=0)

We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).

Then, by definition

P(X=k) = \binom{n}{k}p^k(1-p)^{n-k} = \binom{2}{k}0.4^{k}0.6^{2-k}

where \binom{n}{k}=\frac{n!}{k!(n-k)!}

Then,

P(X=0) = \frac{2!}{0!2!}0.4^{0}0.6^{2} = 0.36

P(X=2)=\frac{2!}{0!2!}0.4^{2}0.6^{0} = 0.16

Then,

E[Y] = 5\cdot 0.16-5\cdot 0.36 = -1

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2 years ago
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