A leather ball has a diameter of 28.4 cm. Which is the area of the piece of leather used on the surface of the ball?
2 answers:
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Answer: The expression is in the explanation.
Step-by-step explanation:
Please find the attached file for the solution.
The answer is 1 gallon.
Miles per gallon(mpg) is computed by dividing the distance traveled by the how many gallons used. So you can derive a formula for how many gallons you would use given the mpg. You will end up with:
The problem asks for how many gallons of gas she will safe in a five-day work work week. So first you need to compute how many miles that would be.
54 miles/day x 5days = 270 miles
So in a five day work week, she will travel 270 miles.
Now to see how much gas she will save, compute how many gallons she will use up for each car, given the mpg of each and find the difference.
First model:30 mpg
This means that with the first model, she will have used up 9 gallons in a 5-day work week.
Second model: 27 mpg
This means that with the second model, she will have used up 10g in a 5-day work week.
Now for the last bit. How much will she save? You can get that by getting the difference of how many gallons each car would have used up.
10gallons - 9gallons = 1g
So she would have saved 1 gallon of gas if she buys the first car instead of the second.
Miles per gallon(mpg) is computed by dividing the distance traveled by the how many gallons used. So you can derive a formula for how many gallons you would use given the mpg. You will end up with:
The problem asks for how many gallons of gas she will safe in a five-day work work week. So first you need to compute how many miles that would be.
54 miles/day x 5days = 270 miles
So in a five day work week, she will travel 270 miles.
Now to see how much gas she will save, compute how many gallons she will use up for each car, given the mpg of each and find the difference.
First model:30 mpg
This means that with the first model, she will have used up 9 gallons in a 5-day work week.
Second model: 27 mpg
This means that with the second model, she will have used up 10g in a 5-day work week.
Now for the last bit. How much will she save? You can get that by getting the difference of how many gallons each car would have used up.
10gallons - 9gallons = 1g
So she would have saved 1 gallon of gas if she buys the first car instead of the second.
Answer:
We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.
Step-by-step explanation:
We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.
A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours
<em>Let </em><em> = population mean amount of time required to produce an electrical part using new procedure</em>
SO, <u>Null Hypothesis</u>, :
45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}
<u>Alternate Hypothesis,</u> :
< 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}
The test statistics that will be used here is <u>One-sample t test statistics </u>because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of time = 43.118 hours
s = sample standard deviation = 5.5 hours
n = sample of parts = 25
So, <u><em>test statistics</em></u> = ~
= -1.711
<em>Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>
Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.