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3 years ago

FIND THE DIFFERENCE

2 answers:

4 0

(5x² + 2x + 11) - (7 + 4x - 2x²)
=5x² + 2x + 11 - 7 - 4x +2x²
=5x²+2x² + 2x-4x + 11-7
=3x²-2x+4

B. 3x2 - 2x + 4

Nadya [2.5K]3 years ago

3 0

Answer:

D. 7x2 - 2x + 4

Step-by-step explanation:

(5x2 + 2x + 11) - (7 + 4x - 2x2)

7x2-2x+4

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Dennis got 85 percent of the questions on the math test correct, which was out of 40. what was his actual mark?

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Brainliest?!?!

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3 years ago

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3 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, we find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

$\sigma = \sqrt{484} = 22$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$5 = 1.96*\frac{22}{\sqrt{n}}$

$5\sqrt{n} = 43.12$

$\sqrt{n} = \frac{43.12}{5}$

$\sqrt{n} = 8.624$

$(\sqrt{n})^{2} = (8.624)^{2}$

$n = 74.4$

We need a sample size of at least 75.

6 0

3 years ago

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