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katovenus [111]
3 years ago
6

HELP ASAP!! PLS HELP IM BEGGING YOU PLS PLS PLS!! (BRIANLIST)

Mathematics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

not sure but I think it is no lines of symmetry.

Step-by-step explanation:

the background with trees when folded together don't line up. they aren't similar.

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1, 2, 3, 6, 103, 206, 309, 618

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Answer:c

Step-by-step explanation: The party will be grand

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Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor
jonny [76]

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5 0
3 years ago
Evaluate 3x² - 1 when x = 2.<br> A. 11<br> B. 35<br> C. 3<br> D. 5
Alexus [3.1K]

Answer:

35

Step-by-step explanation:

All you need to do is substitute!

(3x2)^2-1

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36-1

= 35

Hope this helps!!

4 0
2 years ago
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
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