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Lena [83]
3 years ago
11

A student wants to know how far above the ground the top of a leaning flagpole is. At high​ noon, when the sun is almost directl

y​ overhead, the shadow cast by the pole is 6 ft long. The student holds a plumb bob with a string 2 ft long up to the flagpole and determines that the point of the plumb bob touches the ground 11 in. from the base of the flagpole. How far above the ground is the top of the​ pole?
Mathematics
1 answer:
stepladder [879]3 years ago
4 0

Answer:

idk

Step-by-step explanation:

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3 years ago
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.

3 0
3 years ago
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