A film starts at 15:35 and lasts 132 minutes. What time will it finish

kiruha [24]

Not really, If you use the Pythagorean theorem on all problems related to distance.. You won't always be able to solve it. It depends on the numbers.

So, false.

5 0

3 years ago

Micheal wants to buy 12 pens but he is short of $8.40. If he buys 9 pens, he will have $29.40 left. How much money does he have?

trapecia [35]

All you have to do is subtract the biggest numbers i the problems.

6 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

Scarly bought 3 pounds of cashews. Each pound of cashews cost $2.98. She paid with a $20 bill. How much change did Scarly get ba

pantera1 [17]

Answer:

Part a) Mr. Paul bought 14 oz of cashews

Part b) Mr. Paul bought 12 oz of walnuts

Part c) Mr Paul bought 2 oz more of cashews than walnuts

Part d) Mr Paul bought 2 more ounces of nuts than Mrs Toombs

Step-by-step explanation:

Part a) How many ounces of cashews did Mr. Paul buy?

Remember that

To convert lb to oz multiply by 16

we know that

Mr. Paul bought 7/8 lb of cashews <em> I think hope this helps:)</em>

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Can you please help

Free_Kalibri [48]

Answer:

The answer to the question provided is

$\frac{9}{5}$

Step-by-step explanation:

T

4 0

3 years ago