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Anit [1.1K]
3 years ago
7

Calvin packs boxes of greeting cards into a larger box. The volume of each box of greeting cards is 90 cubic inches. what is the

approximate volume of the large box?

Mathematics
1 answer:
kaheart [24]3 years ago
4 0
Well if you count there are 24 greeting card boxes and the voulme of each is 90 in^3 so it's 2160 in^3.
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g A sample of 3 items is chosen at random from a box containing 20 items, out of which 4 are defective. Let X be the number of d
professor190 [17]

Answer:

The variance and standard deviation of <em>X</em> are 0.48 and 0.693 respectively.

The variance and standard deviation of (20 - <em>X</em>) are 0.48 and 0.693 respectively.

Step-by-step explanation:

The variable <em>X</em> is defined as, <em>X</em> = number of defective items in the sample.

In a sample of 20 items there are 4 defective items.

The probability of selecting a defective item is:

P (X)=\frac{4}{20}=0.20

A random sample of <em>n</em> = 3 items are selected at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 3 and <em>p </em>= 0.20.

The variance of a Binomial distribution is:

V(X)=np(1-p)

Compute the variance of <em>X</em> as follows:

V(X)=np(1-p)=3\times0.20\times(1-0.20)=0.48

Compute the standard deviation (σ (X)) as follows:

\sigma (X)=\sqrt{V(X)}=\sqrt{0.48}=0.693

Thus, the variance and standard deviation of <em>X</em> are 0.48 and 0.693 respectively.

Now compute the variance of (20 - X) as follows:

V(20-X)=V(20)+V(X)-2Cov(20,X)=0+0.48-0=0.48

Compute the standard deviation of (20 - X) as follows:

\sigma (20-X)=\sqrt{V(20-X)} =\sqrt{0.48}0.693

Thus, the variance and standard deviation of (20 - <em>X</em>) are 0.48 and 0.693 respectively.

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Answer:

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