Y= 2 cos(0-pie/6) +2

a bastetball is passed through the air. The height h of the ball in feet after the distance d in feet the ball travels horizonta

Ksenya-84 [330]

You want the (positive) distance when the height is 0, 0= -d^2 + 10d +5. You would factor that out(You'd get radicals), and the final answer would be 5±30−−√ You'd take the only positive one of the two, 5 + srt(30) I hope I did that right :/

3 0

4 years ago

Jeanne traced her coffee mug on her piece of paper. how might she use paper-folding to find the center of the circle?

Leni [432]

If the mug was traced in the center of the paper, she could fold the paper in half vertically and horizontally to find the center point of the mug.

7 0

3 years ago

Read 2 more answers

Solve the system of linear equations using matrices. x+6y=7 x + 5y=4

Rzqust [24]

Answer:

1+6y= "7y=7"

1+5y= "6y=4"

Step-by-step explanation:

X represents "1" so you replace your "X" with "1" and solve the equation.

3 0

3 years ago

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s

user100 [1]

Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

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 $1 - \alpha$ .

That 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.

$M = 2.575\frac{1.7}{\sqrt{36}} = 0.73$

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

4 0

3 years ago

At the Science Museum, there is an exhibition with 150 displays about energy.

tatuchka [14]

1/5

This is a matter of elimination. So let's figure out how many of each display there is. 30% of 150 = 0.30 * 150 = 45 displays about electricity. 2/5 * 150 = 0.40 * 150 = 60 displays about renewable energy 1/10 * 150 = 150/10 = 15 displays about gas. The rest of the displays are about nuclear power. So 150 - 45 - 60 - 15 = 30 displays about nuclear power. Since we want a fraction, we get 30/150 But 30/150 isn't in it's simplest form. The largest number that will evenly divide both 30 and 150 is 30. So divide both the numerator and denominator by 30 (30/30)/(150/30) = 1/5 So 1/5 of the displays are about nuclear power.

8 0

3 years ago