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2 years ago

Lillian has a bag that contains pineapple chews, cherry chews, and peach chews. She

Mathematics

1 answer:

BabaBlast [244]2 years ago

6 0

Answer:

77%

Step-by-step explanation:

since its asking about pineapple and peaches all you have to do is add pineapple and peaches which will give you 36 and since I added all of the numbers together it will give you 36/47

after that you have to put 36/47 in a decimal

36/47 will give you a decimal of .765957 then you got to multiple that by 100 which will give you 76. many numbers all u got to do is round that up to 77 and that's how u get ur answer

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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

5 0

2 years ago

Use the distributive property to solve 4 × 7.1

sladkih [1.3K]

Answer:

28.4

Step-by-step explanation:

To use the distributive property, split the 7.1 to 7 and .1

4*7 would be 28 and 4*.1 would be .4

Adding them together would make the answer 28.4

3 0

3 years ago

Which expression uses the associative property to make it easier to evaluate? (50points)

Naily [24]

The correct answer is option 4: $(8.\frac{1}{4}).\frac{3}{5}$

Step-by-step explanation:

Given expression is:

$8.(\frac{1}{4}.\frac{3}{5})$

The associative property states that for any three terms that are being multiplied, different order of multiplication (i.e. which two terms are multiplied first) will give the same answer.

Let a,b and c be the three terms

then

$a.(b.c) = (a.b).c$