Can someone Please help me find the area of the square I already know the perimeter.

William has 8 1/4 cups of fruit juice. If he divides the juice into 3/4 cup servings, how many servings will he have?

nikdorinn [45]

Answer: 11 servings

Step-by-step explanation:

To get the number of servings, all we need to do is divide 8 1/4 by 3/4

converting 8 1/4 to improper fraction will yield 33/4

8 1/4 ÷ 3/4

=33/4 ÷ 3/4

= 33/4 ÷ 4/3

The 4 at the numerator will cancel the 4 at the denominator. we will be left with;

=33/3 =11

Therefore, there will be 11 servings.

7 0

2 years ago

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Someone answer right por favor

MAXImum [283]

The answer is 64!

hope this helps youu!!

7 0

1 year ago

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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

2 years ago

What is algebraic expression for the difference of 2 and the product of 3 and k

krek1111 [17]

I'm sure the correct answer is 3k-2

5 0

2 years ago

A balloon has a circumference of 28 in. Use the circumference to approximate the surface area of the balloon to the nearest squa

Soloha48 [4]

C=2πr

r=C/(2π), now the surface area is:

A=4πr^2, using r found above we have:

A=4πC^2/(4π^2)

A=C^2/π, we are given that C=28 so:

A=28^2/π in^2

A=784/π in^2

A≈250 in^2 (to nearest square inch)

5 0

2 years ago

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