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2 years ago

How do i do this? can someone help me please

Mathematics

1 answer:

natima [27]2 years ago

8 0

Answer:

i dont know if this is right but the first one is 32.01

Step-by-step explanation:

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What is one three fourths minus seven tenths

Butoxors [25]

1.05, or 21/20

More detailed:

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3 years ago

What is the third quartile of these numbers?

irinina [24]

Sadly the answer has to be at least 20 characters long so I have to say this so I can give you the answer which is 14

8 0

3 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
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 grocery store clerk put only cans of tomato soup and cans of chicken soup on a shelf. The ratio of the number of cans of chick

larisa [96]

Answer:

9 cans of tomato soup

Step-by-step explanation:

We are told that only cans of tomato soup and cans of chicken soup are put on the shelf.

Now, we are told that the ratio of the number of cans of chicken noodle soup to the total number of cans on the shelf was 8 to 17.

In fraction, this is 8/17

Now, this implies that there are 8 cans of chicken noodle soup while there are a total of 17 cans on the shelf.

Thus,

Since only cans of chicken noodles and cans of tomato soup are on the shelf, it means;

Cans of tomato soup = 17 - 8 = 9

8 0

3 years ago

There are 4 teams in a basketball league. Each team plays each of the other teams three times. How many games are played?

svetlana [45]

12 each that is the answer

8 0

4 years ago

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