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3 years ago

Lanas avarage step length is about 0.5 meters. How many steps will she need to finish the race? (5 KILMEMETER RACE)

Mathematics

1 answer:

Alex Ar [27]3 years ago

4 0

The length of the race is 5 kilometers.
5 kilometers equals 5000 meters.

Since Lana's avarege step length is about 0.5 meters, we need to divide 5000 by 0.5

$Steps = \frac{5000}{0.5} = 10000$

Hence, Lana needs to take about 10000 steps.

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Simplify these fraction 9 1/3-5 3/4

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3 years ago

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Which results only in a horizontal compression of y= 1/х by a factor of 6?

marta [7]

Answer:

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3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

Người ta thống kê được rằng mỗi chuyến bay có chừng 0,5% hành khách bị mất hành lí và giá trị trung bình mà khách đòi bồi thường

Butoxors [25]

Answer:

tăng lên 5.000 nghìn đồng

Step-by-step explanation:

mỗi chuyến bay có 0.5% hành khách mất hành lí

mỗi người đòi bồi thường 1.000.000 đồng

do đó vẽ mỗi người tăng lên :0.5%×1.000.000=5.000

5 0

3 years ago

I need to find the greatest common factor of <img src="https://tex.z-dn.net/?f=40s%5E3t%5E5" id="TexFormula1" title="40s^3t^5" a

solong [7]

Answer:

$\large\boxed{GCF(40s^3t^5,\ 2s^4t^4,\ 14s^6t^6)=2s^3t^4}$

Step-by-step explanation:

$40s^3t^5=\boxed2\cdot2\cdot2\cdot5\cdot \boxed{s\cdot s\cdot s}\cdot \boxed{t\cdot t\cdot t\cdot t}\cdot t\\\\2s^4t^4=\boxed2\cdot \boxed{s\cdot s\cdot s}\cdot s\cdot \boxed{t\cdot t\cdot t\cdot t}\\\\14s^6t^6=\boxed2\cdot7\cdot \boxed{s\cdot s\cdot s}\cdot s\cdot s\cdot s\cdot \boxed{t\cdot t\cdot t\cdot t}\cdot t\cdot t\\\\GCF(40s^3t^5,\ 2s^4t^4,\ 14s^6t^6)=\boxed2\cdot \boxed{s\cdot s\cdot s}\cdot\boxed{t\cdot t\cdot t\cdot t}=2s^3t^4$

7 0

4 years ago