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3 years ago

14

Richelle drew a hexagon KLMNOP at the right. She thinks the hexagon has six congruent angles. How can she show that the angles a

re congruent without using a protractor to measure them?

1 answer:

Ipatiy [6.2K]3 years ago

5 0

By using a ruler? If all six sides are the exact same size, they have all same angles.

Hope you get it!

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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

4 years ago

Solve the system of linear equations below.

vekshin1

Answer:

the answer is b. i think so.

5 0

3 years ago

What is the product? Express the answer in lowest terms. 2Two and three-fourths times one-half One and one-eighth One and three-

attashe74 [19]

Answer:

the answer for 2 3/4 times 1 1/8 is 1 and 3/8. The answer for 1 1/8 times 1 3/8 is 6 3/16

Step-by-step explanation:

8 0

3 years ago

Hi! Easy question here, just confused. How do I make the graph? I already created the table, but can you give an example if you

Mademuasel [1]

Answer: Label the axis with numbers (x: from -10 to +10, y from 0 to 50)

Step-by-step explanation: Enter a dot for each coordinate.

7 0

3 years ago

Ramiro drew the graph below.

cluponka [151]

The equation x squared minus 8x+16 represents the data in the graph because if you use the -b/2a x=4. Then if you plug in the 4 for the x in the equation you would come out with the vertex of (4,0). Hope this helps.

4 0

3 years ago

Read 2 more answers