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vodka [1.7K]
3 years ago
11

Where's a numerator found?

Mathematics
2 answers:
Sonja [21]3 years ago
4 0
In a fraction the numerator is the top of the fraction. 

\sf\frac{numerator}{denominator}
lions [1.4K]3 years ago
3 0
A numerator is the number above a line of a common fraction.

For example: In 2/3, the numerator is 2.
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95% of what number is 19?
Setler79 [48]
95% of 20 is 19

Convert your percentage into a decimal: 
\frac{95}{100} = 0.95

Divide 19 by your decimal:
19 \div 0.95 = 20
8 0
3 years ago
Read 2 more answers
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
add 7. double the result. subtract 8. divide by 2. su tract the orginial selected number 1st nuber is 3 second number is 4 the t
sertanlavr [38]
Hi,


The equation looks like this...

n + 7 \times 2 - 8 \div 2 \\ 3 + 7 \times 2 - 8 \div 2 = 3 + 14 - 4 = 13 \\ 4 + 7 \times 2 - 8 \div 2 = 4 + 14 - 4 = 14 \\ 9 + 7 \times 2 - 8 \div 2 = 9 + 14 - 4 = 19 \\ 12 + 7 \times 2 - 8 \div 2 = 12 + 14 - 4 = 22

Hope this helps.
r3t40
8 0
3 years ago
.52 is 10 times as great as what number
WARRIOR [948]

Answer:

is it's right answer is 572

6 0
3 years ago
Math matrix with variable
masya89 [10]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA S

]AS]

L.ASDAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAA

A

AAAAAAA

A

]]]]]AASSSSSSSSSSSSSSAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

8 0
3 years ago
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