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ziro4ka [17]
3 years ago
8

What is 4,321,109,432 rounded to the nearest ten million

Mathematics
2 answers:
Anarel [89]3 years ago
8 0
 4,320,000,000.
Reason:
1. First fine the ten million place.....
2. Look at the next number to the right (3)....
3. If the number is greater than 5 you round up one which is not so you dont round.....
5. Which keeps you at 4,320,000,000.....
If you need anymore help feel free to ask me!
Hope this helps!
Brook aka SlayQueen~
SOVA2 [1]3 years ago
5 0
To round to the nearest ten million, you look at the millions digit. In this case, the millions digit is 1, which is less than 5, so you round the number down to 4,320,000,000.
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(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...

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

The expected value of a Binomial random variable is:

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

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

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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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