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2 years ago

Find the value of the combination. 10C0

Mathematics

2 answers:

Andrej [43]2 years ago

8 0

Answer:The formula for a combination of choosing r unique ways from n possibilities is:

nCr = n!

r!(n - r)!

where n is the number of items and r is the unique arrangements.

Plugging in our numbers of n = 10 and r = 0, we get

10C0 = 10!

0!(10 - 0)!

10C0 = 3,628,800

3,628,800

Vanyuwa [196]2 years ago

6 0

Answer:

Step-by-step explanation:

10C0 = 1.

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Some parts of California are particularly earthquake- prone. Suppose that in one metropolitan area, 25% of all homeowners are in

aleksandrvk [35]

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a. Binomial random variable (n=4, p=0.25)

b. Attached.

c. X=1

Step-by-step explanation:

This can be modeled as a binomial random variable, with parameters n=4 (size of the sample) and p=0.25 (proportion of homeowners that are insured against earthquake damage).

a. The probability that X=k homeowners, from the sample of 4, have eartquake insurance is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{4}{k} 0.25^{0}\cdot0.75^{4}$

The sample space for X is {0,1,2,3,4}

The associated probabilties are:

$P(x=0) = \dbinom{4}{0} p^{0}(1-p)^{4}=1*1*0.3164=0.3164\\\\\\P(x=1) = \dbinom{4}{1} p^{1}(1-p)^{3}=4*0.25*0.4219=0.4219\\\\\\P(x=2) = \dbinom{4}{2} p^{2}(1-p)^{2}=6*0.0625*0.5625=0.2109\\\\\\P(x=3) = \dbinom{4}{3} p^{3}(1-p)^{1}=4*0.0156*0.75=0.0469\\\\\\P(x=4) = \dbinom{4}{4} p^{4}(1-p)^{0}=1*0.0039*1=0.0039\\\\\\$