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2 years ago

Divide;11dq^4/8d^3q^4÷5d/10d^2q^3

Mathematics

1 answer:

Blababa [14]2 years ago

5 0

Answer: 664546^74767^-6382639r575

Step-by-step explanation:

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1174 divided by 45 long division

nekit [7.7K]

Answer:

52,965

Step-by-step explanation:

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3 years ago

Which expression is quick to

Len [333]

Answer:

4⁰ is the answer

Step-by-step explanation:

because first you change the division sign to a multiplication sign because they are the same and then multiply the terms with same base by adding their exponent which looks like this 4⁵-⁷+²

then calculate the sum or difference and you get 4⁰ [hopefully this makes sense]

3 0

3 years ago

Please help with this question

ELEN [110]

Answer:

r^2 * 1/s^4 * t^5

Step-by-step explanation:

r^2 * 1/s^4 * t^5

s^-4 = 1/s^4

7 0

2 years ago

What is the volume of this model

saul85 [17]

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2 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