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miv72 [106K]
3 years ago
9

Find the area of the circle. Use 3.14 for pi. 24 ft

Mathematics
1 answer:
olga_2 [115]3 years ago
8 0
75.36 maybe?? 3.14 x 24
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Eva8 [605]
1. C-7
2. D-4
3. D-16

8. c=7

Im not sure on the rest of those though

6 0
3 years ago
Off Topic: when you delete a place on Life360, does the other person in your circle get the notification that you have remove it
nlexa [21]

Answer:

don't think so

Step-by-step explanation:

8 0
3 years ago
A 109 ft tree casts a shadow that is 130 ft long. what is the angle of elevation of the sun?
Anuta_ua [19.1K]
Answer: 40°


Explanation:

\tan(\theta) = \cfrac{109}{130}

\theta = 40 \textdegree


The Wise Orange wonders about the purpose of knowing the degree of elevation. 
4 0
3 years ago
Please answer this correctly
BabaBlast [244]

Answer:

19.63

Step-by-step explanation:

If the radius is 5 centimeters, then the area is just 3.14*5^2 / 4, which is 78.5/4. This is equal to 19.625. Rounding this give us 19.63 as the final answer

8 0
3 years ago
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
3 years ago
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