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almond37 [142]
3 years ago
8

HELP MOI. Ty friends

Mathematics
2 answers:
Allushta [10]3 years ago
7 0

Answer: B is your answer

Step-by-step explanation: First multiply 5*.19 and you should get 15.95

Next, you should add that additional item that costed 2.95 and you should end up with the equation 15.95 +2.95 and you should get 18.90 as your final answer.

Therefore, $3.1 9/ b is your answer.

*Brainliest answer this plz

stiv31 [10]3 years ago
6 0
$3.19

That will be the answer sir/ma’am
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Step-by-step explanation:

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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
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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
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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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