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3 years ago

What is the answer, why and how?

Mathematics

2 answers:

Elina [12.6K]3 years ago

7 0

The answer to your problem is x^2 + 5x -24

Murljashka [212]3 years ago

7 0

S^2+5s-24 . just multiply

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What is the probability of rolling a number divisible by 3 on a 6-sided standard number cube?

lisabon 2012 [21]

Answer:

Step-by-step explanation:

one out of 3 because there are 6 sides and 1 out of 6 chances getting 3

8 0

3 years ago

Express 236.7 miles in 4.5 days as a unit rate

Andre45 [30]

The answer would be 52.6 miles per 1 day. Take 236.7 and divide it by 4.5.

6 0

4 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

4 years ago

Milan is saving money to buy a game. So far he has saved $20, which is four-fifths of the total cost of the game. How much does

barxatty [35]

Answer:the answer is 16

Step-by-step explanation:

:) hi

8 0

3 years ago

Read 2 more answers

There are twice as many children as adults attending a family reunion. Which fraction of the family members at the reunion are a

pashok25 [27]

Answer:

1/3

Step-by-step explanation:

let a = adults

so then 2a = children

a/2a(+a) = a/3a

6 0

3 years ago