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Zigmanuir [339]
2 years ago
15

Given g(x) = x²+12, find g(8)

Mathematics
2 answers:
vovangra [49]2 years ago
6 0

Answer:

76

Step-by-step explanation:

Given,

g(x) = x^{2} +12

Now,

g(8)=8^{2} +12

     =64 + 12

     = 76

Nikolay [14]2 years ago
5 0

Answer:

76

Step-by-step explanation:

(8)²+12=76

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Step-by-step explanation:

because self is yourself and esteem his how highly you think so how highly you think about your self

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Factor the polynomial by its greatest common monomial factor.<br> 3b^5+15b^4−18b^7
Natasha2012 [34]

Answer:

Step-by-step explanation:

3b⁵ + 15b⁴ - 18b⁷ = 3b⁵ + [3*5*b⁴] - [6*3*b⁷]

=3b⁴*( b + 5 - 6b³)

3 0
3 years ago
There are 12 boys and 24 girls in a sixth grade class, What are 2 statements to compare the number of boys to girls.
Norma-Jean [14]

Answer:

1.) There are twice as many girls as there are boys in a sixth grade class.

2.) There are half as many boys as there are girls in a sixth grade class.

7 0
3 years ago
The student council had a canned food drive. they collected a total of 63 cans on wednesday. this is 21% of the total. how many
Mama L [17]
The total is 300 cans. And 46percent of the total is 138. Therefore 138canned food was collected on Thursday.
4 0
4 years ago
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0
3 years ago
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