A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. the material for the sides of the can costs
0.04 cents per square centimeter. the material for the top and bottom of the can need to be thicker, and costs 0.05 cents per square centimeter. find the dimensions for the can that will minimize production cost.
1 answer:
If you solve this problem generically, you find that the cost of the lateral area of the can will end up being double the cost of the end area of the can.* The ratio of lateral area to end area is h/r so this relation tells you
(0.04h)/(0.05r) = 2
h = 2.5r
Then the radius can be found from the volume equation
V = πr^2*h = 2.5πr^3 = 500 cm^3
r = ∛(200/π) cm ≈ 3.993 cm
h = 2.5r ≈ 9.982 cm
The can is about 4 cm in radius and 10 cm high.
_____
* In the case of a rectilinear shape, the costs of pairs of opposite sides (left-right, front-back, top-bottom) are all the same in the optimum-cost design. It is not too much of a stretch to consider the lateral area of the cylinder to be the sum of left-right and front-back areas, hence twice the cost of the top-bottom area.
(0.04h)/(0.05r) = 2
h = 2.5r
Then the radius can be found from the volume equation
V = πr^2*h = 2.5πr^3 = 500 cm^3
r = ∛(200/π) cm ≈ 3.993 cm
h = 2.5r ≈ 9.982 cm
The can is about 4 cm in radius and 10 cm high.
_____
* In the case of a rectilinear shape, the costs of pairs of opposite sides (left-right, front-back, top-bottom) are all the same in the optimum-cost design. It is not too much of a stretch to consider the lateral area of the cylinder to be the sum of left-right and front-back areas, hence twice the cost of the top-bottom area.
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Answer:
a) Left-skewed
b) We should expect most students to have scored above 70.
c) The scores are skewed, so we cannot calculate any probability for a single student.
d) 0.08% probability that the average score for a random sample of 40 students is above 75
e) If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.
Step-by-step explanation:
To solve this question, we need to understand skewness,the normal probability distribution and the central limit theorem.
Skewness:
To undertand skewness, it is important to understand the concept of the median.
The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.
If the median is larger than the mean, the distribution is left-skewed.
If the mean is larger than the median, the distribution is right skewed.
Normal probabilty distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation, also called standard error of the mean
(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?
Mean = 70, median = 74. So the distribution is left-skewed.
(b) Would you expect most students to have scored above or below 70 points?
70 is below the median, which is 74.
50% score above the median, and 50% below. So 50% score above 74.
This means that we should expect most students to have scored above 70.
(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?
The scores are skewed, so we cannot calculate any probability for a single student.
(d) What is the probability that the average score for a random sample of 40 students is above 75?
Now we can apply the central limit theorem.
This probability is 1 subtracted by the pvalue of Z when X = 75. So
By the Central limit theorem
has a pvalue of 0.9992
1 - 0.9992 = 0.0008
0.08% probability that the average score for a random sample of 40 students is above 75
(e) How would cutting the sample size in half aect the standard error of the mean?
n = 40
n = 20
If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.
Answer:
The answer is
Step-by-step explanation:
To find an equation of a line given the slope and a point we use the formula
where
m is the slope
( x1 , y1) is the point
From the question the point is (−5, 6) and slope = 3
The equation of the line is
We have the final answer as
Hope this helps you