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meriva
3 years ago
13

If Lebron can run 12 laps in 2 hours, approximately how long will it take him to run 60 laps?

Mathematics
1 answer:
ddd [48]3 years ago
3 0

Answer:

10 hours

Step-by-step explanation:

Let's set up a proportion using the following setup.

hours/laps=hours/laps

We know that it takes him 2 hours to run 12 laps. We don't know how long it will take him to run 60 laps, so we can say it takes x hours to run 60 laps.

2 hours/12 laps=x hours/60 laps

2/12=x/60

We want to find out what x is. In order to do this, we have to get x by itself. Perform the opposite of what is being done to the equation. Keep in mind, everything done to one side, has to be done to the other.

x is being divided by 60. The opposite of division is multiplication, so multiply both sides by 60.

60*2/12=x/60*60

60*2/12=x

10=x

It will take Lebron 10 hours to run 60 laps.

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Each year about 1500 students take the introductory statistics course at a large university. This year scores on the nal exam ar
nikitadnepr [17]

Answer:

a) Left-skewed

b) We should expect most students to have scored above 70.

c) The scores are skewed, so we cannot calculate any probability for a single student.

d) 0.08% probability that the average score for a random sample of 40 students is above 75

e) If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

Step-by-step explanation:

To solve this question, we need to understand skewness,the normal probability distribution and the central limit theorem.

Skewness:

To undertand skewness, it is important to understand the concept of the median.

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

If the median is larger than the mean, the distribution is left-skewed.

If the mean is larger than the median, the distribution is right skewed.

Normal probabilty distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation, also called standard error of the mean s = \frac{\sigma}{\sqrt{n}}

(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?

Mean = 70, median = 74. So the distribution is left-skewed.

(b) Would you expect most students to have scored above or below 70 points?

70 is below the median, which is 74.

50% score above the median, and 50% below. So 50% score above 74.

This means that we should expect most students to have scored above 70.

(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?

The scores are skewed, so we cannot calculate any probability for a single student.

(d) What is the probability that the average score for a random sample of 40 students is above 75?

Now we can apply the central limit theorem.

\mu = 70, \sigma = 10, n = 40, s = \frac{10}{\sqrt{40}} = 1.58

This probability is 1 subtracted by the pvalue of Z when X = 75. So

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{75 - 70}{1.58}

Z = 3.16

Z = 3.16 has a pvalue of 0.9992

1 - 0.9992 = 0.0008

0.08% probability that the average score for a random sample of 40 students is above 75

(e) How would cutting the sample size in half aect the standard error of the mean?

n = 40

s =  \frac{10}{\sqrt{40}} = 1.58

n = 20

s =  \frac{10}{\sqrt{20}} = 2.24

If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

4 0
2 years ago
During a recent rainstorm, 888 centimeters of rain fell in Dubaku's hometown, and 2.362.362, point, 36 centimeters of rain fell
Elenna [48]

Answer:

<em>5.64 centimeters more rain fell in Dubaku's town than in Elliot's town.</em>

Step-by-step explanation:

Amount of rainfall in Dubaku's hometown is 8 centimeters, amount of rainfall in Elliot's hometown is 2.36 centimeters and the amount of rainfall in Alperen's hometown is 15.19 centimeters.

So, <u>the difference of amount of rainfall for Dubaku's town and Elliot's town</u> will be:  (8- 2.36) centimeters = 5.64 centimeters.

That means, 5.64 centimeters more rain fell in Dubaku's town than in Elliot's town.


7 0
3 years ago
Read 2 more answers
Janice Corrudo is paid a commission of 15% of her yearly sales up to $85,000 and 18% of any sales above $85,000. Her total sales
LekaFEV [45]

Answer:

Idk lol

Step-by-step explanation:

4 0
2 years ago
Please help xxxxxxxx
Free_Kalibri [48]

Answer:

  • a. \bold{45000\times \:10^7}
  • b. \bold{3500}

Step-by-step explanation:

a. \left(5\times \:10^3\right)\left(9\times \:10^7\right)

5\times 10^3 = 5000\\\\5000\times \:9\times \:10^7

\mathrm{Multiply\:the\:numbers:}\:5000\times \:9=45000

=45000\times \:10^7

b. \left(7\times 10^5\right)\div \left(2\times 10^2\right)

\left(7\times 10^5\right)\div \left(2\times 10^2\right)=\frac{7\times \:10^5}{2\times \:10^2}

\frac{10^5}{10^2} = 10^3

\frac{7\times \:10^3}{2}

\mathrm{Factor}\:\: 10^3 : 2^3\times5^3

\frac{7\times \:2^3\times \:5^3}{2}

\frac{2^3}{2}=2^2

7\times \:2^2\times \:5^3=3500

7 0
2 years ago
BRAINLY AND 100 POINTSIn the figure below, angle y and angle x form vertical angles. Angle x forms a straight line with the 50°
melisa1 [442]
40+50=90 all the sides are equally split so y = 90
6 0
3 years ago
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