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3 years ago

Solve the equation for the red variable(C)in R-C=P

Mathematics

1 answer:

BARSIC [14]3 years ago

5 0

If you want to solve the equation

$R-C=P$

for C, you can work like this: subtract P from both sides:

$R-C-P = 0$

Add C to both sides:

$R-P=C$

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What algebraic expression must be added to the sum of 3x2 +4x+8 and 2x2−6x+3 to give 9x2−2x−5 as the result?

vesna_86 [32]

Answer:

$4x^2-16$

Step-by-step explanation:

-This is an addition/subtraction problem.

#We subtract the sum of the first two expression from the third expression:

$\#Add \ first \ two\\\\=3x^2 +4x+8 +2x^2-6x+3\\\\=5x^2-2x+11\\\\\# Subtract\ from \ the \ third\\=(9x^2-2x-5)-(5x^2-2x+11)\\\\=9x^2-2x-5-5x^2+2x-11\\\\=4x^2-16$

Hence, the expression $4x^2-16$ must be added to the first two.

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3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago