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OLga [1]
3 years ago
12

The weight of an apple is 150g rounded to the nearest 5g

Mathematics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

The weight of an apple is 150g rounded to the nearest 5g

what is the lower bound weight of the apple

what is the upper bound of the weight of the apple

Step-by-step explanation:

The weight of an apple is 150g rounded to the nearest 5g

what is the lower bound weight of the apple

what is the upper bound of the weight of the apple

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Which one is question 20??
8 0
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I will mark brainliest if correct please answer fast I will give 10 points to the first person who answers
attashe74 [19]

Answer:

99\ in^{2}

Step-by-step explanation:

Area of any triangle could be always calculated as base length multiplied by the height perpendicular to the aforementioned base divided by 2. Therefore:

A = 18 * 11 / 2 = 198 / 2 = 99

4 0
3 years ago
If m 1 = 40, find the measure of 7
TEA [102]
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5 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
The following two points are on a line: (2,3), (-2,5). What is the slope of the line?
8_murik_8 [283]

Answer:

D

Step-by-step explanation:

slope = rise / run

slope = (5 - 3) / (-2 - 2)

slope = 2 / -4

slope = -2 / 4

slope = -1/2

8 0
1 year ago
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