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4 years ago

Can someone help me with number 135?

Mathematics

1 answer:

Nat2105 [25]4 years ago

3 0

so from accidental deaths we have

30 from falls

47 from fires

200 from drowning

1950 car accidents

so the total of those is 30+47+200+1950, or 2227.

the ones that weren't car accidents will be 2227 - 1950, or 277.

if we take 2227 to be the 100%, how much is 277 off of it in percentage?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 2227&100\\ 277&x \end{array}\implies \cfrac{2227}{277}=\cfrac{100}{x}\implies x=\cfrac{277\cdot 100}{2227} \\\\\\ x\approx 12.4382577~\hspace{7em}\stackrel{rounded~up}{x=12}$

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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.

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Answer:

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Step-by-step explanation:

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