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zysi [14]
3 years ago
14

Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

Mathematics
1 answer:
jekas [21]3 years ago
6 0

Answer:

Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16% Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%

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Please help!
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Answer:

The zeros are:

x =4, x=2, x = 5

  • The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

0=\left(x-4\right)^2\left(x^2-7x+10\right)

switch sides

\left(x-4\right)^2\left(x^2-7x+10\right)=0

as

x^2-7x+\:10=\left(x-2\right)\left(x-5\right)

so

\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0

Using the zero factor principle

  • \mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

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Thus, the zeros are:

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It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

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Hence, option (B) is true.

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If he eats more then 12 lunches in a month he should take Plan B, if he eats less then 12 lunches a month he should take Plan A.

Step-by-step explanation:

2.5x12=30

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Ken borrowed $ 2000 from Sam at 8% per year.what is the simple interest after 6 years? A.$ 160 B.$ 960 C. $ 1,600 D. $ 9,600
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Step-by-step explanation:

2000×8/100=160

160×6=960

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