All categories

3 years ago

Can you PLEASE HELP ME!!!

Mathematics

2 answers:

Nuetrik [128]3 years ago

8 0

Answer:

The correct answer would be B I believe

Step-by-step explanation:

ss7ja [257]3 years ago

3 0

The correct answer is d

You might be interested in

More math help please

Vladimir [108]

C. it is an or problem so it goes below and it does not have the equal to so it is a dashed line.

7 0

4 years ago

Factor 2x 4 - 20x 2 - 78.

const2013 [10]

2(x^4 - 10x^2 - 39)
2(x^2 - 13)(x^2 + 3)

Letter B

6 0

3 years ago

Read 2 more answers

Find (f×g)(x) when f(x)=x^2+8x+15 and g(x)=5 over x^2-9

Kipish [7]

Look at the picture......

5 0

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

4 years ago

Read 2 more answers

I need help finding the solutions

maks197457 [2]

I believe that x = 2 and y = -2

4 0

3 years ago

Read 2 more answers