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Ne4ueva [31]
3 years ago
14

Two sides of a triangle measure 9 cm and 23 cm. Which could be the

Mathematics
2 answers:
vampirchik [111]3 years ago
7 0
It would be 148cm because 9+23=32 and subtract 32 from 180 would get you 148
gavmur [86]3 years ago
5 0

Answer:

h²=b²+p²

x²= 9²+23²

x²= 81+529

<em>x²</em><em>=</em><em>6</em><em>1</em><em>0</em>

<em>X=</em><em> </em><em>√</em><em>6</em><em>1</em><em>0</em>

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Which method would determine the volume of the prism with dimensions 2 times 2 and one-fourth times 4 shown below?
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Step-by-step explanation:

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Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins
Lina20 [59]

Answer:

(a) The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b) The most likely value for <em>X</em> is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number among the four homeowners  who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.

The random sample of homeowners selected is, <em>n</em> = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.

The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.33\\=1.32

Thus, the most likely value for <em>X</em> is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

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3 years ago
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Masja [62]

Answer:

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Step-by-step explanation:

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