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3 years ago

HELLPPP DUE TODAY AT 4

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Answer:

2 and 2/3 for the first question

-1/8 for the second one

-1/5 for the third one

ive never done this type of question or maybe it was too long ago for me to remember but out of this answer choices i would guess that its the last one

this is 3

Step-by-step explanation:

you add the numerators which is the top number in order to get 2 and 2/3rds

you subtract 5 from 4 to get -1/8

same situation as the last one, subtract 15 from 11 to make it a negative 4/20 and then simplify that by dividing it by four to get 1/5th

this one i dont understand or remember how to do very well but if i had to guess its the last answer choice

this one you add 3/4 plus 1/4 in order to get 4/4 or rather 1, so basically its asking what 1+1+1 is which is 3

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2 years ago

10. A random sample of 10 fibers is taken from a collection of 92 fibers that consists of 43 fibers of polymer A, 17 fibers of p

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Answer:

0.1150 = 11.50% probability that the sample does not contain any fibers of polymer B

Step-by-step explanation:

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fibers are chosen without replacement, which means that we use the hypergeometric distribution.

92 fibers means that $N = 92$

Sample of 10 means that $n = 10$

No polymeter B which means that $x = 0$

17 fibers of polymeter B which means that $k = 17$

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This is P(X = 0).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,92,10,17) = \frac{C_{17,0}*C_{75,10}}{C_{92,10}} = 0.1150$

0.1150 = 11.50% probability that the sample does not contain any fibers of polymer B

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