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asambeis [7]
3 years ago
8

I'm pretty sure the answer is median, just need confirmation, has a picture. Please help! Thanks!!

Mathematics
2 answers:
anygoal [31]3 years ago
8 0
It's a median you are right
nikdorinn [45]3 years ago
4 0
I think it is an angle bisector. 
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Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi
BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

\mu = 6*0.6*3 + 6*0.4*1 = 13.2

The standard deviation is:

\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

\mu = 6*0.6 = 3.6

The probability of scoring four goals is:

P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122

Probability of two assists:

40% of 2 are assists, which means that:

\mu = 6*0.4 = 2.4

The probability of getting two assists is:

P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127

Probability of four goals and two assists:

P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05

0.05 = 5% probability that he has four goals and two assists in one game

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What is the simplest ratio of 32 and 48
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The graph of f(x) shown below has the same shape as the graph of g (x) = x^2 but it is shifted to the left 2 units. what is the
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Answer:

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