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3 years ago

Help can someone help me find the value of ClX Y and Z I would appreciate it luvvs ;) ♡

Mathematics

1 answer:

Stolb23 [73]3 years ago

6 0

$\\ \sf\longmapsto 9x+46=35$

$\\ \sf\longmapsto 9x=35-46$

$\\ \sf\longmapsto 9x=-11$

$\\ \sf\longmapsto x=\dfrac{-11}{9}$

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3 years ago

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Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,

Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.

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Answer:

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Help can someone help me find the value of ClX Y and Z I would appreciate it luvvs ;) ♡​

Help can someone help me find the value of ClX Y and Z I would appreciate it luvvs ;) ♡