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Yuki888 [10]
3 years ago
14

Solve the formula for C. Profit: P=R−C C=

Mathematics
1 answer:
Kipish [7]3 years ago
6 0
You want to isolate C, so the first step would to have it be positive. You would do this by adding C to both sides of the equation to cancel the C on one of the sides out like this:
p + c = r - c + c \\ p + c = r
. Then you want to cancel out the P, so you subtract it on both sides like this:
c + p - p = r - p \\ c = r - p
And that is your answer
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Find the length of the segment with endpoints of (3,2) and (-3,-6).
MakcuM [25]
The formula of the length of the segment AB:
A(x_A;\ y_A);\ B(x_B;\ y_B)\\\\|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}
We have:
A(3;\ 2)\to x_A=3;\ y_A=2\\\\B(-3;\ -6)\to x_B=-3;\ y_B=-6
Substitute:
|AB|=\sqrt{(-3-3)^2+(-6-2)^2}=\sqrt{(-6)^2+(-8)^2}\\\\=\sqrt{36+64}=\sqrt{100}=10
Answer: B) 10.

5 0
3 years ago
Rearrange the equation A = xy to solve for x.
mote1985 [20]
X = A/y
You divide the y from both sides
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5 0
2 years ago
Find the solution to the system:<br>( 6(x +y)-y--1<br>| 7(y +4)–(y + 2)=0<br><br><br>plssss help​
siniylev [52]

Answer:

the answer to the questions are x, 4 and y=7

4 0
3 years ago
What are the zeros of f(x)=(x+5)(x-9)?
klemol [59]

Answer:

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Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
2 years ago
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