A 4-foot long steel pipe consists of two concentric cylinders, with the inner cylinder hollowed
out. The radius of the outside of the pipe is 6 inches and the radius of the inside of the pipe
is 5.75 inches.
HINT: The units of measure must be the same! Convert to inches and keep your answer in
terms of π.
A. Determine the volume of metal used to build the pipe.
B. If the pipe is to be powder-coated on the inside and outside surfaces, what is the total
surface area to be powder-coated?
Answers
Answer:
The pipe is formed by two concentric cylinders.The outside cylinder has 6 inches of radius.The inside cylinder has 5.75 inches of radius.To find the volume of the pipe, we need to subtract the inside cylinder volume from the outside cylinder volume.
Remember that the volume of a circular cylinder is
[tex]V=\pi r^{2} h[/tex]
Where [tex]r[/tex] is the radius and [tex]h[/tex] is the height.
Outside cylinder volume.[tex]V_{outside}=\pi r^{2}h= \pi (6in)^{2} (48in)=1,728 \pi in^{3}[/tex]
Inside cylinder volume.[tex]V_{inside}=\pi r^{2}h= \pi (5.75in)^{2} (48in)=1,587 \pi in^{3}[/tex]
Notice that we used the height 4 feet in inches units, that's why the height in the formulas is 48 inches, because each feet is equivalent to 12 inches.
Volume of the pipe.[tex]V_{pipe}=V_{outside} -V_{inside} =1,728 \pi in^{3}-1,587 \pi in^{3} =141 \pi in^{3}[/tex]
(A) Therefore, the volume of metal used to build the pipe is 141π cubic inches.
Now, to know the amount of powder-coat we must use, we need to find the surface area of the pipe, which is basically the sum of the surface area of both cylinders.
Surface area of outside cylinder.[tex]S_{outside}=2\pi r^{2}+2\pi rh=2 \pi (6in)^{2}+2 \pi (6in)(48in)= 72 \pi in^{2} +576 \pi in^{2} =648 \pi in^{2}[/tex]Surface area of the inside cylinder.[tex]S_{inside}=2\pi r^{2}+2\pi rh=2 \pi (5.75in)^{2}+ 2 \pi (5.75in) (48in)= 66.13 \pi in^{2} +552 \pi in^{2} =618.13 \pi in^{2}[/tex]
The total surface is[tex]S_{powder}=648 \pi in^{2} + 618.13 \pi in^{2} =1,266.13 \pi in^{2}[/tex]
(B) Therefore, we need 1,266.13π sqaure inches of powder to cover the whole pipe.
Answer:
A. volume of the metal used to build the pipe is 141[tex]\pi[/tex] cubic inches.
B. The total surface area to be powder coated is 1122.125 square inches.
Step-by-step explanation:
The length of the cylinder = 4 feet = 48 inches.
Radius of the outside of the pipe = 6 inches.
Radius of the inside of the pipe = 5.75 inches
A. volume of the metal used to build the pipe = volume of the outside pipe - volume of the inside pipe
volume of a cylinder = [tex]\pi[/tex][tex]r^{2}[/tex]h
volume of the outside pipe = [tex]\pi[/tex][tex]r^{2}[/tex]h
= [tex]\pi[/tex] × [tex]6^{2}[/tex] × 48
= 1728[tex]\pi[/tex] cubic inches
volume of the inside pipe = [tex]\pi[/tex][tex]r^{2}[/tex]h
= [tex]\pi[/tex] × [tex]5.75^{2}[/tex] × 48
= 1587[tex]\pi[/tex] cubic inches
volume of the metal used to build the pipe = 1728[tex]\pi[/tex] - 1587[tex]\pi[/tex]
= 141[tex]\pi[/tex] cubic inches
B. Total surface area of a hollow cylinder = 2[tex]\pi[/tex] ( [tex]r_{1}[/tex] + [tex]r_{2}[/tex]) ( [tex]r_{2}[/tex] - [tex]r_{1}[/tex] + h)
where [tex]r_{1}[/tex] is the inner radius and [tex]r_{2}[/tex] is the outer radius.
= 2[tex]\pi[/tex] (6 + 5.75)(5.75 - 6 + 48)
= 2[tex]\pi[/tex] (11.75 × 47.75)
= 1122.125[tex]\pi[/tex] square inches
The total surface area to be powder coated is 1122.125 square inches.
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23. A popular brand of AAA batteries has an effective use time of 12.3 hours, on average. A startup company claims that their AAA batteries last longer. The startup company tested 24,000 of their new batteries and computed a mean effective use time of 12.32 hours. Although the difference is quite small (72 seconds—or just over a minute), the effect was statistically significant (P-value < 0.0001).
It is appropriate to conclude which of the following?
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B) The startup company has moderate evidence that their AAA batteries last longer, on average.
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Answers
Answer:
Option B
Step-by-step explanation:
Although the difference is quite small (72 seconds—or just over a minute), the effect was statistically significant (P-value < 0.0001).
Since the P-value was less than 0.0001, the null hypothesis in this case can be rejected (H₀: ∪=12.3), concluding that the startup company has moderate/enough evidence that their AAA batteries last longer, on average.
Answer:
Step-by-step explanation:
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For the null hypothesis,
µ = 12.3
For the alternative hypothesis,
µ > 12.3
This is a right tailed test.
The decision rule is to accept the null hypothesis if the significance level is lesser than the p value and reject the null hypothesis if the significance level is greater than the p value.
Let us assume a significance level of 0.05.
Since alpha, 0.05 > than the p 0.0001, then we would reject the null hypothesis.
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Answers
Answer:
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Step-by-step explanation:
The volume of a cone is given by the following formula:
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Two cones:
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Skinny:
radius r, height h. So
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Sharing size:
radius 3r, height h. So
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About how many times more popcorn does the sharing cone hold than the skinny cone?
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Step-by-step explanation:
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0
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Step-by-step explanation:
Answer:
About 296 households have more than three children.
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Answers
Answer:
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Step-by-step explanation:
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Answers
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Answers
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12
Step-by-step explanation:
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Answers
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Step-by-step explanation:
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The physical plant at the main campus of a large state university recieves daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 52 and a standard deviation of 5. Using the empirical rule, what is the approximate percentage of lightbulb replacement requests numbering between 47 and 52
Answers
Answer:
34% of lightbulb replacement requests numbering between 47 and 52
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 52
Standard deviation = 5
Between 47 and 52:
52 is the mean and 47 is one standard deviation below the mean.
By the Empirical Rule, 68% of the measures are within 1 standard deviation of the mean.
Since the normal distribution is symmetric, of those, 34% are within 1 standard deviation below the mean and the mean(47 and 52) and 34% are within the mean and one standard deviation above the mean(52 and 57).
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34% of lightbulb replacement requests numbering between 47 and 52
Panat has assets that total $9,500. His liabilities total $1,900. Which expression will find Patrick's net worth?
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$1.900-$9.500
$1.900-$9.500
$9.500-$1.900
Helppp
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Answer:
D
Step-by-step explanation:
assets total - liabilities/deductions = net worth
his income is what his net worth it, but liabilities is what you dont have. its what you owe/need to pay/dont own. subtracting these two will give you your real net wroth.
The correct expression to find Patrick's net worth is D: $9,500 - $1,900.
What is the net worth?The difference between an individual's assets and liabilities is described as net worth.
As per the question, Patrick's assets are given as $9,500 and his liabilities are given as $1,900.
To find his net worth, we need to subtract his liabilities from his assets.
Thus, the expression for Patrick's net worth is:
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Net worth = $9,500 - $1,900
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Therefore, Patrick's net worth is $7,600.
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Answer:
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1hr 2hr 3hr 4hr 30 min
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Step-by-step explanation:
8 x
105 is how many times as great as 8 x 10-l?
10 4
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10 -4
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Answer:
10 6
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The lengths of pregnancies are normally distributed with a mean of 250 days and a standard deviation of 15 days.
a. Find the probability of a pregnancy lasting 308 days or longer?
b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8% (8th percentile), find the length that separates premature babies from those who are not premature.
Answers
Answer:
a) 0.005% probability of a pregnancy lasting 308 days or longer
b) The pregnancy length that separates premature babies from those who are not premature is 229 days.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 250, \sigma = 15[/tex]
a. Find the probability of a pregnancy lasting 308 days or longer?
This is 1 subtracted by the pvalue of Z when X = 308. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{308 - 250}{15}[/tex]
[tex]Z = 3.87[/tex]
[tex]Z = 3.87[/tex] has a pvalue of 0.99995
1 - 0.99995 - 0.00005
0.005% probability of a pregnancy lasting 308 days or longer
b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8% (8th percentile), find the length that separates premature babies from those who are not premature.
The 8th percentile is X when Z has a pvalue of 0.08. So it is X when Z = -1.405.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.405 = \frac{X - 250}{15}[/tex]
[tex]X - 250 = -1.405*15[/tex]
[tex]X = -1.405*15 + 250[/tex]
[tex]X = 229[/tex]
The pregnancy length that separates premature babies from those who are not premature is 229 days.
A pollen grain is 3/10 to the third power centimeters wide. In an illustration, the pollen grain is 6 centimeters 10 wide. How much larger is the illustration than the actual pollen?
Answers
Answer:
The illustration is approximately 226 times larder than the actual pollen grain.
Step-by-step explanation:
First of all, we have to write out the size of the actual pollen grain in a manner that is easy for us to work with.
This will be
[tex](\frac{3}{10}) ^{3}=0.027cm[/tex] wide
The original width of the pollen is 0.027 cm.
Next,
We can take the width of the illustration to be 6.10 cm wide.
After this,
To get the magnification factor, we will have to use the formula
Magnification = illustrated size/ actual size
This will be the magnification = [tex]6.10/0.027[/tex] = 255.92 times
Hence, the illustration is approximately 226 times larder than the actual pollen grain.
