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jek_recluse [69]
3 years ago
10

Find coordinates of point c

Mathematics
1 answer:
BaLLatris [955]3 years ago
4 0
Ans n solution can refer in the picture

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If f(x) = (-x)3, what is f(-2)?
Ilia_Sergeevich [38]

Answer: f(-2) should be 6

Step-by-step explanation:

6 0
3 years ago
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10xy(4xy^3-7xy+9y^2)
Thepotemich [5.8K]

Answer:

40x^{2} y^{4} -70x^{2} y^{2} +90xy^{3}

Step-by-step explanation:

10xy(4xy^{3} -7xy+9y^2)  

1. 10xy · 4xy^3 = 40x^2y^4

2. 10xy · -7xy = -70x^{2} y^{2}

3. 10xy · 9y^2 = 90xy^3

Final step: add up all of those values together to make an equation!

Answer:  40x^2y^4+ -70x^2y^2+ 90xy^3

Final Answer: 40x^2y^4-70x^2y^2+90xy^3

3 0
3 years ago
A​ student's course grade is based on one midterm that counts as 20​% of his final​ grade, one class project that counts as 20​%
Rus_ich [418]

Answer:

His overall final​ score is 80.1.

His letter grade is a B.

Step-by-step explanation:

To find his grade, we multiply each grade by it's weight.

Grades and weights:

His midterm score is 64​. The midterm counts 20% = 0.2.

His project score is 80​. The project score counts 20% = 0.2.

His homework score is 94. The homework score counts 30%.

His final exam score is 77. It counts 30%.

What is his overall final​ score?

64*0.2 + 80*0.2 + 94*0.3 + 77*0.3 = 80.1

His overall final​ score is 80.1.

What letter grade did he earn​ (A, B,​ C, D, or​ F)?

At least 80 but less than 90 is a​ B. He scored 80.1, so his letter grade is a B.

5 0
3 years ago
How many women must be randomly selected to estimate the mean weight of women in one age group.we want 90% confidence that the s
mojhsa [17]
That is a very good question....

Look Grank

you know that:

confidence interval = mean +or- Margin of Error

and

Margin of Error = (z)*(standard deviation) / (sqrt of n)
where n is the number of sample records

So we need to calculate z-value firstly

It says: "<span>we want 90% confidence"

So that:
</span>
<span>confidence90% corresponds to z-value of 1.645

Plug that into our formula of </span><span>Margin of Error:

</span>Margin of Error = (1.645)*(22) / (sqrt of n)

"The sample mean is within 2.7 lb of the population mean" means that Margin of Error is 2.7

<span>Margin of Error:

</span>2.7 = (1.645)*(22) / (sqrt of n)

Now solve for n:

n=179.66~180

SO that 180<span> women must be randomly selected to estimate the mean weight of women in one age group.</span>
7 0
3 years ago
PLEASE HELP ME!!!
miskamm [114]
The answer is:

\frac{3y(2x-1)(x-3}{x(x-6)(x+1)}
4 0
3 years ago
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