She walks 2 minutes for every 5 she runs. If she walks for 12 minutes how many minutes does she spend running?

1 answer:

5 0

Answer:

I believe it is, 30 minutes of running.

Explanation:

If she spends 12 minutes walking, that means she’s done 6 ‘sets’ of two minutes.

Then, you just have to multiply 6 by 5 and then you get 30.

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A bucket holds 10 potatoes. Hal removes and cleans 4 potatoes per minute.

nexus9112 [7]

Answer:

an = 10 – 4(n – 1)

Step-by-step explaination

answer is an = 10 – 4(n – 1)

4 0

2 years ago

3) What is the greatest common factor of 5x7, 30x4, and 10x3?

saveliy_v [14]

30 is the greatest common factor I think

6 0

3 years ago

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In a box of 250 paperclips, 50 are red. What percent of paperclips are red

nadezda [96]

50/250 = 1/5 = 0.20 = 20%

Answer: 20%

8 0

3 years ago

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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

What would be the equation for #23???

Rufina [12.5K]

Hi there!

Let r be the number of matches that Rebecca's team won.

We know that Katheryn's team won 9 more matches than Rebecca's team, and we know that Katheryn's team won 52 matches.

Using this given information, we can create the following equation.

$r+9=52$

If you wish to solve this, we can subtract both sides by 9 to get the following value for r.

$r=43$

Have an awesome day! :)

~collinjun0827, Junior Moderator

8 0

2 years ago

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