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3 years ago

Ok so do I subtract 5 from x and 5 Orr am I solving wrong i don’t know this is confusing.

Mathematics

1 answer:

Aleksandr [31]3 years ago

4 0

Answer:

no you dont

Step-by-step explanation:

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The quality assurance engineer of a receiving-sets manufacturer inspects receiving-sets in lots of 50. He selects 4 of the 50 re

Mariana [72]

Answer:

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

Step-by-step explanation:

Sets are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes 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 successes.

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:

Lots of 50 means that $N = 50$

5 are defective, which means that $k = 5$

4 are selected, which means that $n = 4$

Find the probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

This is $P(X = 2)$ . So

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

$P(X = 2) = h(2,50,4,5) = \frac{C_{5,2}*C_{45,2}}{C_{50,4}} = 0.0430$

0.0430 = 4.30% probability that exactly 2 of the 4 receiving-sets selected by the engineer are defective.

4 0

3 years ago

Find the length of each model. Then find the scale factor. (Pictures attached #3/#4)

max2010maxim [7]

12 cm for the first one and 11.6 for the second.

Unrelated to the question, that house around number 3 gives me nostalgia of elementary school :)

6 0

3 years ago

Pls help Eithan needs to divide theese base ten blocks (9 ten blocks and 6 one blocks)

sineoko [7]

Answer:

carrot

Step-by-step explanation:

give me brainlyist i never got it pls bt i have snap if you want im a girl

7 0

3 years ago

The time period of a simple pendulum is directly related to the square root of its length. An 8 foot pendulum has a period of 3.

svp [43]

$T_1=k \sqrt{8}$

$T_2=k \sqrt{3}$

$\frac{T_2}{T_1} = \frac{T_2}{3.2} = \frac{ \sqrt{3} }{ \sqrt{8} }$

Then:

$T_2=3.2 \sqrt{ \frac{3}{8} } =1.96$ secs

6 0

3 years ago

Hi, I really need help with this ASAP!

Talja [164]

Answer:

google maybe

Step-by-step explanation:

3 0

3 years ago