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3 years ago

Please help! If it's not any of the multiple choice please tell me

Mathematics

1 answer:

Vesnalui [34]3 years ago

6 0

Its not any of them. The answer is 48. She did it completely wrong
-4/9(-9/4)=x/100(-9/4)

the next step would be:
-1=-9x/432
Then the answer would be 48 since you would multiply by 432 and divide by -9

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Larry says 5×900 is less than 5×9000 is greater than 9000

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Answer:

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Step-by-step explanation:

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2 years ago

Which property of inequality should you use to solve 3x ≤ 27?

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2 years ago

Two equations are given below:

ch4aika [34]

Rewrite the equations;
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3 years ago

The mass of 2 large boxes is 16,000 g. One box has a mass of 9,000 g. What is the mass of the other box in kilograms? (1,000 g =

allochka39001 [22]

Answer:

7 kg

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3 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