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3 years ago

Can someone help me with all of this plz

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

Here's a solution for #8
Volume of a sphere = V = 4/3π r³
Your radius r = 6, so plugging that in you get:
V = 4/3π(6³)
V = 4/3π216
V = 864/3π
V = 288π (units³)

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3 years ago

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With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the en

mrs_skeptik [129]

Answer:

0.9167 = 91.67% probability that the entire batch will be accepted

Step-by-step explanation:

The CDs are chosen without replacement, which means that the hypergeometric distribution is used.

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.

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

2100 write-rewrite CDs, and 60 are defective.

This means, respectively, that $N = 2100, k = 60$

3 of these CDs are randomly selected for testing

This means that $n = 3$

What is the probability that the entire batch will be accepted?

Probability of none defective, so P(X = 0).

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

$P(X = 0) = h(0,2100,3,60) = \frac{C_{60,0}*C_{2040,3}}{C_{2100,3}} = 0.9167$

0.9167 = 91.67% probability that the entire batch will be accepted

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2 years ago