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nignag [31]
2 years ago
5

The weights of NCAA lacrosse players are approximately normally distributed with a mean of 182 lbs and a standard deviation of 1

3.1 lbs.
. a) Find the probability that a player will have a weight between 163 lbs and 191 lbs. (6 points)

. b) How much would a lacrosse player weigh if he was lighter than 75% of the other players? (5 points)

Show your work in proper format on separate paper making sure you've answered all parts of the question.

Then, snap a pic of your work and upload it as your answer.

Be sure your upload is the correct file.
Mathematics
1 answer:
Marizza181 [45]2 years ago
4 0

Answer:

What can I do for you

Step-by-step explanation:

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Answer:

length of 1 side of A, using the Pyth. Thm. and the dimensions of the other two squares: (side of A)^2 = (10 in)^2 + (24 in)^2. Then:

(side of A)^2 = 100+ 576 in^2 = 676 in^2.

Here I have not bothered to solve for the length of the side of A, since we want the area of square A. But if you do want the side length, find it: sqrt(676) = 26 in. Then the area of A is (26 in)^2 = 676 in^2.

Then the area of square A is (26 in)^2 =

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Step-by-step explanation:

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3 years ago
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A pool pumps fills 1/6 of the pool in 1/3 hour. How much of the swimming pool is fillied in one hour? Can u please explain and c
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Step-by-step explanation:

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2 years ago
A rare form of malignant tumor occurs in 11 children in a​ million, so its probability is 0.000011. Four cases of this tumor occ
Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

\lambda=r*t=0.000011*13,280=0.14608

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990

c) The probability of more than one case is:

P(k>1)=1-P(k\leq 1)=1-0.990=0.010

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0
3 years ago
Jack wants to find out which can of pineapples is the best buy. He has a choice of four different-sized cans. Which can has the
OLga [1]

Answer:

C

Step-by-step explanation:

Here is the complete question

ack wants to fined out which can pineapples is the best buy. He has a choice of four different -sized cans. Which can has the lowest unit price per ounce?

A. 8 ounce for $ 1.44

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unit price = price / total ounces

$1.44 / 8 = $0.18

$1.14 / 6 = $0.19

$2.88 / 18 = $0.16

$1.62 / 10 = 0.162

the lowest is 18 ounce for $ 2.88

5 0
3 years ago
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