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bogdanovich [222]
3 years ago
6

A 2-column table with 7 rows. Column 1 is labeled sample with entries 1, 2, 3, 4, 5, 6, 7. Column 2 is labeled Sample Mean with

entries 154, 121, 160, 135, 140, 134, 129. Looking at this table of sample means, which value is the best estimate of the mean of the population?
Mathematics
2 answers:
grandymaker [24]3 years ago
8 0

Answer:

(B)136

Step-by-step explanation:

Rom4ik [11]3 years ago
7 0

Answer:

(B)136

Step-by-step explanation:

Edg 2020 your welcome luv <3

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The area of a trapezoid is 864cm². It has a height of 24 cm, and the length of one of its bases is 30 cm. What is the length of
Amiraneli [1.4K]

The area <em>A</em> of a trapezoid with height <em>h</em> and bases <em>b</em>₁ and <em>b</em>₂ is equal to the average of the bases times the height:

<em>A</em> = (<em>b</em>₁ + <em>b</em>₂) <em>h</em> / 2

We're given <em>A</em> = 864, <em>h</em> = 24, and one of the bases has length 30, so

864 = (<em>b</em>₁ + 30) 24 / 2

864 = (<em>b</em>₁ + 30) 12

864 = (<em>b</em>₁ + 30) 12

72 = <em>b</em>₁ + 30

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When adding two<br> rational numbers what<br> will the sign be if both<br> numbers are positive?
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Step-by-step explanation:

Adding two rational numbers results in a positive sum.

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PLEASE ANSWER ASAP! PLEASE SHOW ALL WORK!
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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
3 years ago
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