A yogurt costs 55p. How many yogurts can be bought for three pounds?

write the equation of a line in slope intercept form that is parallel to the line x+5y=10 and goes through the point (-5,6)

lilavasa [31]

Answer:

The equation would be y = -1/5x + 5

Step-by-step explanation:

To find the equation of this new line, we first need to identify the slope of the first line. We do this by putting it into slope intercept form.

x + 5y = 10

5y = -x + 10

y = -1/5x + 2

In slope-intercept form, the slope is always the coefficient of x. This makes the slope of the first equation -1/5. Since parallel lines have the same slope, our new slope will also be -1/5.

Given that and the point, we can solve for the equation using point-slope form.

y - y1 = m(x - x1)

y - 6 = -1/5(x + 5)

y - 6 = -1/5x - 1

y = -1/5x + 5

4 0

3 years ago

Find the variable of x. 8x-2=-9+7x

TiliK225 [7]

Answer:

8x-2=-9+7x equals X=-7

7 0

3 years ago

Read 2 more answers

You have one type of nut that sells for $4.50/lb and another type of nut that sells for $8.00/lb. You would like to have 28 lbs

-Dominant- [34]

Answer:

I will need 26.4 pounds of the expensive nuts and 1.6 pounds of the cheap nuts.

Step-by-step explanation:

Since I have one type of nut that sells for $ 4.50 / lb and another type of nut that sells for $ 8.00 / lb, and I would like to have 28 lbs of a nut mixture that sells for $ 7.80 / lb, to determine how much of each nut will I need to obtain the desired mixture, the following calculation must be performed:

8 x 0.95 + 4.5 x 0.05 = 7.825

8 x 0.94 + 4.5 x 0.06 = 7.79

0.94 x 28 = 26.32

26.4 x 8 + 1.6 x 4.5 = 218.4

218.4 / 28 = 7.8

Thus, I will need 26.4 pounds of the expensive nuts and 1.6 pounds of the cheap nuts.

7 0

3 years ago

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

So, the 95% confidence interval is (46.4, 53.6)

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

So, the 95% confidence interval is (47.9, 52.1)

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

Notice that the margin of error decreases when the sample size increases.

4 0

3 years ago

What is the volume of the pyramid in the diagram?

Jlenok [28]

The volume of a pyramid is lwh/3

4 0

3 years ago