SOMEONE ANSWER THIS QUICK!!

What is the greatest common factor of 42 and 90 and 81

o-na [289]

3 is the GCF of them

4 0

3 years ago

Read 2 more answers

Stem (hundred thousands) Leaf (ten thousands)

alina1380 [7]

The largest stem is 3, found in the bottom row. This represents 300 thousand or 300,000.

The largest leaf in the bottom row is 7 and that represents

7*(10 thousand) = 70 thousand = 70,000

Adding the stem and leaf values gets us

300,000 + 70,000 = 370,000

<h3>The most expensive home sold was $370,000</h3>

6 0

3 years ago

HELP PLEASE!!<br><br> Which one is correct? =/ or = ?

Delicious77 [7]

The answer is = since each equation equals 32
6•6=36
2•2=4
36-4=32
And then the next equation
6+2=8
4(8)=32

3 0

1 year ago

Read 2 more answers

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

A feeder at the zoo in shape of a cone has a radius of 3 inches. It holds about 113.04 cubic inches of food. Approximate its hei

umka2103 [35]

Answer:

12 inches

Step-by-step explanation:

3 0

3 years ago