All categories

1 year ago

I dont really kno what to do.

Mathematics

1 answer:

ASHA 777 [7]1 year ago

4 0

Answer:

Step-by-step explanation:

rearrange the equation to the letter on the left side until you find an equation equal to one of the answers

You might be interested in

Which equation below represents a rate of change of -3?

Amanda [17]

The answer is C

Is shows a change of rate for the integer

Hope this helps

6 0

2 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

Elodia [21]

Answer:

is this geomtey

Step-by-step explanation:

Make sure to put what kind of maths (geometry algebra ect.) this is so others can meet your needs better. Also might want to put what unit or type of problem this is as well.

5 0

2 years ago

The Question

77julia77 [94]

Answer:

Step-by-step explanation:

Divide 12 by .40 (the percentage) to figure out what the full number would be.

5 0

2 years ago

A=pi r^2 which if the following equations can be used to solve for the radius (r)

zhannawk [14.2K]

C. You would first have to divide A by Pi, then take the square root of r, in order to get r by itself.

4 0

2 years ago

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma

IrinaVladis [17]

Answer:

538 books should be tested.

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$ , so $z = 2.575$

Now, 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.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when $M = 0.08$

We have that $\sigma = 0.72$

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

$0.08 = 2.575*\frac{0.72}{\sqrt{n}}$

$0.08\sqrt{n} = 2.575*0.72$

$\sqrt{n} = \frac{2.575*0.72}{0.08}$

$(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}$

$n = 537.1$

Rounding up

538 books should be tested.

6 0

2 years ago

I dont really kno what to do.​

I dont really kno what to do.