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3 years ago

What is the prime factorization of 72?

Mathematics

2 answers:

yawa3891 [41]3 years ago

8 0

Prime factorization can be obtained by the factor tree
72=2*36=2*(2*18)=2*(2*(2*9))=2*2*2*3*3=2^3*3^2

GREYUIT [131]3 years ago

8 0

Answer:

Just to keep it simple-the correct answer is D.

Step-by-step explanation:

2^3 x 3^2=

(2 x 2 x 2) x (3 x 3)=

8 x 9 = 72

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2 years ago

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
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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.

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