All categories

4 years ago

12 minutes to drive 30 laps; 48 minutes to drive 120 laps

1 answer:

7 0

You can divide 12min.➗30lps. Then you divide 48➗120lps.
And the answer will be 12➗30=0.4 then 48➗120=0.4 and the answer is that the ratios or rates are equvalent by dividing.

You might be interested in

Divide (16x6 − 12x4 + 4x2) by 4x2.

Degger [83]

We know that
(16x6 − 12x4 + 4x2) / 4x2
is equal to
[16x6/4x2]+[-12x4/4x2]+[4x2/4x2]
=[4x4]+[-3x2]+[1]
=4x4-3x2+1

the answer is
4x4-3x2+1

6 0

3 years ago

2(5x+4)=10x+6 Someone can help me?

Oliga [24]

Answer:

4 0

3 years ago

Read 2 more answers

Please round to the nearest tenth if possible.

Butoxors [25]

$\bf \cfrac{64~\underline{mi}}{\underline{hr}}\cdot \cfrac{\underline{hr}}{60~sec}\cdot \cfrac{5200~ft}{\underline{mi}}\implies \cfrac{64\cdot 5200~ft}{60~sec}\implies \cfrac{16640~ft}{3~sec}
\\\\\\
\approx 5546.67\frac{ft}{sec}\\\\
-------------------------------\\\\
\cfrac{6000~\underline{lbs}}{\underline{day}}\cdot \cfrac{ton}{2000~\underline{lbs}}\cdot \cfrac{7~\underline{day}}{week}\implies \cfrac{6000\cdot 7~ton}{2000~week}\implies 21.00\frac{ton}{week}$

$\bf -------------------------------\\\\
\cfrac{2~\underline{lbs}}{\underline{week}}\cdot \cfrac{16~oz}{\underline{lbs}}\cdot \cfrac{\underline{week}}{7~day}\implies \cfrac{2\cdot 16~oz}{7~day}\implies \cfrac{32~oz}{7~day}\approx 4.57\frac{oz}{day}$

8 0

4 years ago

Amanda started the week with $200 in her bank account. Then she deposited $300 she earned from a garage sale. What was the final

Doss [256]

500 dollars 300+200=500

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