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3 years ago

Vupti vvo 1.5(A): Solving Linear Equatior your work. 4n-10 =12

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

n = 5.5

Step-by-step explanation:

4n - 10 = 12

4n = 12 + 10

4n = 22

n = 22/4

n = 5.5

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Convert 3 quarts to pints

ss7ja [257]

Answer:

6 pints

Step-by-step explanation:

There are two pints in a quart, so there are 2 times 3 pints in three quarts.

Hope this helps!

4 0

3 years ago

Read 2 more answers

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

3 years ago

Debra has two-thirds of a pound of loose green tea leaves. If she wants to keep one-half of the tea at home and the other half a

MrMuchimi

Answer: <em>I would personally say 1/3</em>

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2 years ago

What are the coordinates of the endpoints of the midsegment for △JKL that is parallel to KL¯¯¯¯¯? Enter your answers in the boxe

labwork [276]

Answer:

J(-2, -1), K(4, -5), L(0, -5)

The midpoint of segment JL is

(-2 + 0)/2, (-1 + (-5))/2) = (-2/2, -6/2) = (-1, -3)

The midpoint is

(-2 + 4)/2, (-1 + (-5))/2) = (2/2, -6/2) = (1, -3)

A: (-1, -3), (1, -3)

Step-by-step explanation:

5 0

3 years ago

The length of a rectangle is 6 inches more than the width. the perimeter is 28 inches. find the length and width.

GalinKa [24]

The solution of your problem is shown on the picture.

7 0

3 years ago