Answer:
From a statistical point of view: A correct decision was made.
so that the extent of the problem may be ascertained.
Step-by-step explanation:
"In statistical hypothesis testing, a type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion), while a type II error is the non-rejection of a false null hypothesis (also known as a "false negative" finding or conclusion)."
Final answer:
The scenario described indicates that a Type I error was made during the hypothesis test, as the machine was working correctly but was stopped due to the rejection of the null hypothesis.
Explanation:
The question asks about the consequences of a hypothesis test related to the functioning of a machine filling quart size bottles of olive oil. When the hypothesis is rejected and the machine is shut down, but no issue is found upon inspection, this scenario describes a Type I error. This error occurs because the null hypothesis, which would be the assumption that the machine is functioning correctly, is rejected even though it is actually true.
A biologist observes that a certain bacterial colony triples every 4 hours and after 12 hours occupies 1 square centimeter. Assume that the colony obeys the population growth law. The area the colony occupied when first observed was:___________ (a) 1/9 sq. cm (b) 1/81 sq. cm. (c) 1/36 sq. cm (d) 1/27 sq. cm. (e) None of the above.
Answer:
DC C C C C C C C C C CSDC SCS DC SDCS DC SDC S SDC SDC SDCS DC SDC SDC DV DFV DV DFV F CSDC SD CS DCS DC SDC SD C SDDVSVV FV FDV DFV D VDFC C C C C C C C DSCQ 1 1 1 1W13 3
Step-by-step explanation:
Final answer:
To determine the original area of the bacterial colony, the final size of 1 square centimeter is divided by the growth factor of 3 for each 4-hour interval, which results in 1/27 square cm.
Explanation:
The question concerns the concept of exponential growth seen in populations, specifically in a bacterial colony. We are given that the bacterial colony triples every 4 hours and that after 12 hours the colony occupies 1 square centimeter. To find the area the colony occupied when first observed, we need to calculate the size of the colony 12 hours ago.
To solve this, we divide the final size by the growth factor for each interval that has passed.
1 square cm (size after 12 hours) / 3 (growth after 4 hours) = 1/3 square cm after 8 hours
1/3 square cm / 3 (another 4 hours growth) = 1/9 square cm after 4 hours
1/9 square cm / 3 (another 4 hours growth) = 1/27 square cm at time first observed
Three people have been nominated for president of a class. From a poll, it is estimated that the first candidate has a 37% chance of winning and the second candidate has a 44% chance of winning. What is the probability that the third candidate will win?
Answer:
19%
Step-by-step explanation:
We assume that the class has to have a president. So the chances of none of the candidates winning is 0%. In turn we expect that one of the candidates will win the elections, therefore the chances of someone winning is 100%. Therefore the chances of the third candidate winning can be calculated by removing the chances of the other two candidates winning
P(third candidate winning) = 100% - (37%+44%) = 19%
On a game show, you are given five digits to arrange in the proper order to form the price of a car. If you are correct, then you win the car. What is the probability of winning, given the following conditions?
(a) You guess the position of each digit.
(b) You know the first digit and guess the positions of the other digits.
Answer:
(a) 0.00833
(b) 0.04167
Step-by-step explanation:
There are 5 pieces to form a car.
Total number of arrangement of these 5 pieces is, [tex]5!=5\times4\times3\times2\times1 = 120[/tex]
Of these 120 arrangements only 1 arrangement will form a proper car.
(a)
Probability that each position's guess is correct is,
[tex]P(Winning)=\frac{Favorable\ arrangements}{Total\ number\ of\ arrangements} \\=\frac{1}{120}\\ =0.00833\\\approx0.833\%[/tex]
Thus, the probability of getting all the guesses correct is 0.00833 or 0.833%.
(b)
It is given that we know the first correct piece.
That is we need to guess the other 4 from the 4 remaining pieces.
Total number of arrangement of these 5 pieces is,
[tex]4!=4\times3\times2\times1 = 24[/tex]
Of these 24 arrangements only 1 arrangement will form a correct arrangement with the known first piece.
Probability that each position's guess is correct is,
[tex]P(Winning)=\frac{Favorable\ arrangements}{Total\ number\ of\ arrangements} \\=\frac{1}{24}\\ =0.04167\\\approx4.17\%[/tex]
Thus, the probability of getting all the guesses correct when we know the first correct piece is 0.04167 or 4.17%.
The probability of winning when
(a) You guess the position of each digit is 1/120(b) You know the first digit and guess the positions of the other digits is 1/24How to determine the probabilities?The number of digits is given as:
n = 5
When you guess the position of each digit, the number of combination is:
n! = 5!
Expand
n! = 5 * 4 * 3 * 2 * 1
n! = 120
Only one of the 120 combinations is right.
So, the probability of winning is 1/120
When you guess the position of other four digits, the number of combination is:
n! = 1 * 4!
Expand
n! = 1 * 4 * 3 * 2 * 1
n! = 24
Only one of the 24 combinations is right.
So, the probability of winning is 1/24
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. A system contains two components, A and B. The system will function only if both components function. The probability that A functions is 0.98, the probability that B functions is 0.95, and the probability that either A or B functions is 0.99. What is the probability that the system functions?
Answer:
0.94
Step-by-step explanation:
System will function if both components function, so,
P(system function)=P(A∩B)=?
P(A∩B)=P(A)+P(B)-P(A∪B)
We are given that P(A)=0.98, P(B)=0.95 and P(A or B)=P(A∪B)=0.99.
P(A∩B)=0.98+0.95-0.99=1.93-0.99=0.94
P(system function)=P(A∩B)=0.94.
Thus, the probability that the system functions is 0.94 or 94%.
The probability that the system functions, requiring both A and B to function, is calculated using the formula for the probability of either event occurring. By rearranging and substituting the given probabilities, we find that the probability the system functions is 0.94.
Explanation:Calculation of System Functionality ProbabilityTo determine the probability that the system functions, we need to find the joint probability that both A and B function, denoted as P(A AND B). Given that the probability A functions is 0.98, and B functions is 0.95, we use the given that the probability either A or B functions (which includes the case where both function) is 0.99.
We start with the formula for the probability that either A or B functions, which is:
P(A OR B) = P(A) + P(B) − P(A AND B)
.
We can rearrange this to solve for P(A AND B):
P(A AND B) = P(A) + P(B) − P(A OR B)
.
Substituting the given probabilities, we get:
P(A AND B) = 0.98 + 0.95 − 0.99 = 0.94
.
Therefore, the probability that the system functions, which requires both A and B to function, is 0.94.
Nicole breeds pit-bull terriers. Her dog. Bella, gave birth to 7 puppies. For the female pup she charges $550 and $500 for males. Her profit comes out to be $3650, determine how many males and female were born.
Answer:
The answer to your question is 4 males and 3 females were born.
Step-by-step explanation:
Data
total number of puppies = 7
price of females = f = $550
price of males = m = $500
Profit = $3650
Process
1.- Write two equations that represent this problem
f + m = 7
550f + 500m = 3650
2.- Solve this system of equations by substitution
f = 7 - m
550(7 - m) + 500m = 3650
Simplify
3850 - 550m + 500m = 3650
Solve for m
- 550m + 500m = 3650 - 3850
- 50m = -200
m = -200/-50
m = 4
3.- Calculate f
f = 7 - 4
f = 3
After the exam has been completed, you have the students anonymously fill out a questionnaire asking about their study habits for the exam and the grade they earned on the exam. From the surveys, you randomly select 10 students who studied for the exam and 10 students who did not study for the exam.
You create the table showing the students' exam grades given here:
Grades of 10 students who studied Exam grade:
94 96 90 88 88 100 78 95 97 94
Grades of 10 students who did not study Exam grade:
64 73 71 64 56 49 89 67 76 71
What was the average exam grade for each set of students? Enter the average exam grade of students who studied, followed by the average exam grade of the students who did not study, using two significant figures, separated by a comma.
Answer:
Students who studied
[tex]\bar X =\frac{94+96+90+88+88+100+78+95+97+94}{10}=\frac{920}{10}=92[/tex]
Students who did not study
[tex]\bar X =\frac{64+73+71+64+56+49+89+67+76+71}{10}=\frac{680}{10}=68[/tex]
The answer would be:
92,68
Step-by-step explanation:
For this case we have the following data:
Students who studied
Exam grade: 94 96 90 88 88 100 78 95 97 94
The sample mean is calculated with the following formula:
[tex]\bar X= \frac{\sum_{i=1}^{10} X_i}{n}[/tex]
And if we replace the values given we got:
[tex]\bar X =\frac{94+96+90+88+88+100+78+95+97+94}{10}=\frac{920}{10}=92[/tex]
Students who did not study
Exam grade: 64 73 71 64 56 49 89 67 76 71
The sample mean is calculated with the following formula:
[tex]\bar X= \frac{\sum_{i=1}^{10} X_i}{n}[/tex]
And if we replace the values given we got:
[tex]\bar X =\frac{64+73+71+64+56+49+89+67+76+71}{10}=\frac{680}{10}=68[/tex]
You roll the same die three times. Consider the possible outcomes if the order of the resultsis not recorded (meaning, e.g., that 1-2-1 and 2-1-1 are considered the same outcome).(a) Order the possible results in lexicographical order and show by direct counting that thenumber of possible outcomes is
Answer: 56 outcomes
Step-by-step explanation:
Tossing a die 3 times and not having the order recorded, we have the following outcome.
[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]
[1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],
[1,3,3] [1,3,4] [1,3,5] [1,3,6]
[1,4,4] [1,4,5] [1,4,6]
[1,5,5] [1,5,6]
[1,6,6]
[2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],
[2,3,3] [2,3,4] [2,3,5] [2,3,6]
[2,4,4] [2,4,5] [2,4,6]
[2,5,5] [2,5,6]
[2,6,6]
[3,3,3] [3,3,4] [3,3,5] [3,3,6]
[3,4,4] [3,4,5] [3,4,6]
[3,5,5] [3,5,6]
[3,6,6]
[4,4,4] [4,4,5] [4,4,6]
[4,5,5] [4,5,6]
[4,6,6]
[5,5,5] [5,5,6]
[5,6,6]
[6,6,6]
Hence, By direct counting, the number of possible outcome is 56 outcomes.
Consider a data set with at least three data values. Suppose the highest value is increased by 10 and the lowest is decreased by 10. For each of the following, explain by example or use about 2 or 3 sentences.a. Does the mean change? Explain.b. Does the median change? Explain.c. Is it possible for the mode to change? Explain.
Answer:
a. Mean doesn't change.
b. Median doesn't change.
c. Mode can change.
Step-by-step explanation:
Let us assume the data set with 10 observations
{2,6,4,3,2,6,4,9,4,7}.
Arranging data set in ascending order
{2,2,3,4,4,4,6,6,7,9}
mean=2+2+3+4+4+4+6+6+7+9/10=4.7
median
n/2=10/2=5 is an integer so,
median= average of n/2 and n/2 +1
median= (5th value+6th value)/2
median=(4+4)/2=8/2=4
Mode
The most repeated value is 4. So, mode is 4 for assumed data.
Increasing highest value by 10 and decreasing lowest value by 10
{-8,2,3,4,4,4,6,6,7,19}
a.
mean=-8+2+3+4+4+4+6+6+7+19/10=4.7
Mean doesn't change
b.
median
n/2=10/2=5 is an integer so,
median= average of n/2 and n/2 +1
median= (5th value+6th value)/2
median=(4+4)/2=8/2=4
Median doesn't change
c.
Most occurring value is still 4. But mode can change if the value the highest value becomes most concurring value.
A. The mean may or may not change, depending on the original data values. If the data set has an odd number of values and the highest and lowest values are not the same, then the mean will change.
B. The median will not change because it is the middle value of the data set, and adding or subtracting a constant value to all data points does not affect the relative order of the values. The position of the median will remain the same
C. The mode may or may not change. If the mode was the highest or the lowest value in the original data set and its frequency did not change after the adjustments, the mode will remain the same.
A.However, if the data set has an even number of values or the highest and lowest values are the same, then the mean will remain unchanged. For example:
Original data set: 5, 8, 10, 12, 15
After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5
Mean before: (5+8+10+12+15)/5 = 10
Mean after: (25+8+10+12+(-5))/5 = 10
B. For example: Original data set: 5, 8, 10, 12, 15
After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5
Median before: 10 (middle value of the ordered data set)
Median after: 10 (still the middle value of the ordered data set)
C. However, if the highest or lowest value was not the mode in the original data set and its frequency becomes the highest after the adjustments, the mode will change. For example:
Original data set: 5, 8, 10, 12, 15
After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5
Mode before: No mode (all values are unique)
Mode after: 10 (frequency of 10 is now higher than other values)
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100 point question.......
Answer:
4 and 8/21
Step-by-step explanation:
w-5/7=3 2/3
Find 3 2/3 + 5/7
3 2/3 + 5/7 = 3 14/21 + 15/21 = 3 29/21 = 4 8/21
W= 4 8/21
Answer:
w = 92/21
Step-by-step explanation:
Step 1: Convert into improper fractions
5/7 -> 5/7
3 2/3 -> 3 * 3/3 + 2/3 -> 9/3 + 2/3 -> 11/3
Step 2: Make common denominator
5/7 * 3/3 -> 15/21
11/3 * 7/7 -> 77/21
Step 3: Add 15/21 to both sides
w - 15/21 + 15/21 = 77/21 + 15/21
w = 92/21
Answer: w = 92/21
A department store has daily mean sales of $28,372.72. The standard deviation of sales is $2000. On Tuesday, the store sold $34,885.21 worth of goods. Find Tuesday's z-score. Was Tuesday a significantly good day?
Answer:
Tuesday z-score was 3.26.
Tuesday was a significantly good day.
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.
A score is said to be significantly high if it has a z-score higher than 1.64, that is, it is at least in the 95th percentile.
In this problem, we have that:
[tex]\mu = 28372.72, \sigma = 2000[/tex]
On Tuesday, the store sold $34,885.21 worth of goods. Find Tuesday's z-score.
This is Z when [tex]X = 34885.21[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34885.21 - 28372.72}{2000}[/tex]
[tex]Z = 3.26[/tex]
Tuesday z-score was 3.26.
Was Tuesday a significantly good day?
A z-score of 3.26 has a pvalue of 0.9994. So only 1-0.9994 = 0.0006 = 0.06% of the day are better than Tuesday.
So yes, Tuesday was a significantly good day.
Tuesday's z-score is 3.26, which is greater than 2, indicating that Tuesday was a significantly good sales day.
Explanation:The z-score is a measure of how many standard deviations an element is from the mean. To calculate it, we subtract the mean from the amount sold on Tuesday, and then divide by the standard deviation.
The formula for calculating a z-score is: Z = (X - μ) / σ, where X is the value we are looking at (in this case Tuesday's sales), μ is the mean and σ is the standard deviation.
So, Tuesday's z-score would be calculated as follows:
Z = ($34,885.21 - $28,372.72) / $2000 = 3.26
Since the z-score is greater than 2, this is considered statistically significant, and thus would indicate that Tuesday was indeed a significantly good sales day.
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Fill in the blanks. Optionshouse tracked the performance of their most active day traders and found that the probability of a winning call option pick was 0.5375. If in a day, 458 call options are picked by these traders, around __________ of them will be winners, give or take __________. Assume each pick is independent.a. 246.2, 113.8500 b. 246.2, 10.67 c. 10.67, 246.2 458, d. 10.67 246.2, 0.5375
Answer:
If in a day, 458 call options are picked by these traders, around 246.2 of them will be winners, give or take 10.67 .
Step-by-step explanation:
Hello!
Your study variable is X: the number of winning calls in a sample of 458 calls.
The variable has a binomial distribution since you have two possible outcomes, that the call is a winning call (success) or that the call is not a winning call (failure), each call is independent and the probability of success is p= 0.5375 and the probability of failure q= 1-p= 1-0.5375= 0.4625.
The expected value for a binomial distribution is
E(X)= n*p= 458 * 0.5375= 246.175
And to know the standard error (or standard deviation) you have to calculate the square root of the variance:
V(X)= n*p*q= 458*0.5375*0.4625= 113.85
√V(X)= √113.85= 10.67
I hope it helps!
In a small company with 20 employees, 10 employees make $80,000/yr, 6 employees make $ 150,000/yr the 4 highest-paid employees all make $220,000/yr. Calculate the average salary in the company
Answer:
$129 000/yr
Step-by-step explanation:
Weighted average is used to answer this question.
total employees are 20
10 employees make 80 000
total earning for 10 employees = 80 000 * 10 = 800 000 (multiplying)
6 employees make 150 000
total earning for 6 employees= 150 000 * 6 =900 000 (multiplying)
4 employees make 220 000
total earning for 4 employees = 220 000 * 4 = 880 000 (multiplying)
To calculate weighted average all the totals are added and then divide by total number of employees.
weighted average = (800 000 + 900 000 + 880 000)/20
weighted average = 2580000/20
Weighted average = 129 000
A city has streets laid out in a square grid, with each block 135 mm long. If you drive north for three blocks, then west for two blocks, how far are you from your starting point? Express your answer in meters.
Total distance traveled from the starting point = 487 mm (or) 0.487 m
Solution:
Given data:
Length of each block = 135 mm
Distance traveled towards North = three blocks
= 135 × 3
= 405
Distance traveled towards North = 405 mm
Distance traveled towards West = two blocks
= 135 × 2
= 270
Distance traveled towards West = 270 mm
Total distance traveled
[tex]=\sqrt{\text{Distance traveled in North}^2+\text{Distance traveled in West}^2}[/tex]
[tex]=\sqrt{405^2+270^2}[/tex]
[tex]=\sqrt{236925}[/tex]
[tex]=135\sqrt{13}[/tex]
= 487 mm (approximately)
Total distance traveled from the starting point = 487 mm
Let us convert mm to m.
1 m = 1000 mm
487 mm = 487 ÷ 1000 = 0.487 m
Total distance traveled from the starting point = 0.487 m
The distribution of hours of sleep per week night, among college students, is found to be Normally distributed, with a mean of 6.5 hours and a standard deviation of 1 hour. What range contains the middle 95 % of hours slept per week night by college students?
Answer:
4.5 and 8.5 hours.
Step-by-step explanation:
The empirical 95% rule states that, in a normal distribution, 95% of the data falls within two standard deviations bellow or above the mean. If the mean is 6.5 hours and the standard deviation is 1 hour, the interval is:
[tex]6.5-2*1\leq x \leq 6.5+2*1\\4.5\leq x \leq 8.5[/tex]
The range that contains the middle 95 % of hours slept per week night by college students is 4.5 to 8.5 hours.
Using the concept of Normal Distribution, and knowing that 95% of data falls within two standard deviations, we find that hours of sleep among students range between 4.5 and 8.5 hours.
Explanation:This question involves the concept of a Normal distribution in statistics, commonly used in analyzing patterns of data spread. In this scenario, we know the sleep hours are normally distributed with a mean (average) of 6.5 hours and a standard deviation of 1 hour. The middle 95% of hours slept per week night by college students means we are looking for the range of sleep hours that falls within two standard deviations of the mean on a normal distribution curve.
Because the standard deviation is 1 hour, two standard deviations would be 2 hours. Thus, we subtract and add two standard deviations from the mean to find the range. That is, 6.5 - 2 = 4.5 hours (lower bound) and 6.5 + 2 = 8.5 hours (upper bound).
So, the middle 95% of sleep hours per week night by college students falls within the range of 4.5 to 8.5 hours.
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An English class consists of 23 students, and three are to be chosen to give speeches in a school competition. In how many different ways can the teacher choose the team, given the following conditions?
Answer: Check the attached
Step-by-step explanation:
The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off employees at its Sam's Club warehouse unit. Approximately half of the layoffs will be hourly employees (The Wall Street Journal, January 25-26, 2014). Suppose the following data represent the percentage of hourly employees laid off for Sam's Club stores.
55 56 44 43 44 56 60 62 57 45 36 38 50 69 65
a. Compute the mean and median percentage of hourly employees being laid off at these stores. Mean Median
b. Compute the first and third quartiles. First quartile Third quartile
c. Compute the range and interquartile range. Range Interquartile range
d. Compute the variance and standard deviation. Round your answers to four decimal places. Variance Standard deviation
e. Do the data contain any outliers
Answer:
a) [tex] \bar X = 52[/tex]
[tex] Median = 55[/tex]
b) [tex]Q_1= \frac{44+44}{2}=44[/tex]
[tex]Q_3= \frac{57+60}{2}=58.5[/tex]
c) [tex] Range = Max -Min = 69-36=33[/tex]
d) [tex] s^2 =100.1429[/tex]
[tex] s= \sqrt{100.143}=10.0071[/tex]
e) [tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]
[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]
A possible outlier would be the value of 69 since its above the upper limti for the boxplot.
Step-by-step explanation:
For this case we have the following dataset:
55 56 44 43 44 56 60 62 57 45 36 38 50 69 65
A total of 15 observations
Part a
We calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^{15} X_i}{15}[/tex]
And for this case we got [tex] \bar X = 52[/tex]
For the median we ust need to order the data on increasing way like this:
36, 38,43,44,44,45,50,55,56,56,57,60,62,65,69
Since the number of observations is an odd number the median would be on the 8 position from the dataset ordered on this case:
[tex] Median = 55[/tex]
Part b
In order to calculate the Q1 we need to select the following data:
36, 38,43,44,44,45,50,55
And the Q1 would be the average between the 4 and 5 positions like this:
[tex]Q_1= \frac{44+44}{2}=44[/tex]
And for the Q3 we select these values:
55,56,56,57,60,62,65,69
And the Q3 would be the average between the 4 and 5 positions like this:
[tex]Q_3= \frac{57+60}{2}=58.5[/tex]
Part c
The Range is defined as:
[tex] Range = Max -Min = 69-36=33[/tex]
Part d
In order to calculate the sample variance we can use the following formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And if we replace we got:
[tex] s^2 =100.1429[/tex]
And the deviation is just the square root of the variance:
[tex] s= \sqrt{100.143}=10.0071[/tex]
Part e
For this case we need to find the lower and upper limits for the boxplot given by:
[tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]
[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]
A possible outlier would be the value of 69 since its above the upper limti for the boxplot.
Availability is the most important consideration for designing servers, followed closely by scalability and throughput. a. [10]<1.7>Wehaveasingleprocessorwithafailureintime(FIT)of100.What is the mean time to failure (MTTF) for this system
Answer:
a) Mean Time to failure (MTTF) = (10^7) hours
b) Availability of the system = 1
c) Mean Time to failure for 1000 processors = 10^4 hours.
Step-by-step explanation:
a) Failures in time (FIT) is traditionally reported as failure Per billion hours Of Operation.
1 billion = (10^9)
FIT = 100/(10^9) = 10^-7
MTTF = 1/FIT = 1/(10^-7) = (10^7) hours
b) Availability of the system = MTTF/(MTTF + MTTR)
MTTR = mean time to repair = 24hours
Availability of the system = (10^7)/((10^7) + 24) = 0.9999 = 1
c) FIT = 1000 (processors) × 100 (FIT per processor) = (10^5) per billion hours of operations = (10^5)/(10^9) = 10^-4
MTTF = 1/FIT = 1/(10^-4) = (10^4) hours
QED!!
(a) MTTF for a single processor:[tex]\(10^7\)[/tex] hours.
(b)System availability: approximately 0.9999976.
(c) MTTF for a system with 1000 processors: [tex]\(10^4\)[/tex] hours.
Let's address each part of the problem step-by-step, providing detailed explanations and calculations.
Part (a): Mean Time to Failure (MTTF)
The Mean Time to Failure (MTTF) is calculated from the Failures in Time (FIT) rate. FIT is defined as the number of failures per billion [tex](10^9)[/tex] hours of operation.
Given:
- Failures in Time (FIT) = 100
First, convert FIT to the failure rate (λ).
[tex]\[ \lambda = \frac{\text{FIT}}{10^9} \text{ failures per hour} \][/tex]
So,
[tex]\[ \lambda = \frac{100}{10^9} \text{ failures per hour} \][/tex]
The MTTF is the reciprocal of the failure rate (λ):
[tex]\[ \text{MTTF} = \frac{1}{\lambda} \][/tex]
Now, calculate MTTF:
[tex]\[ \text{MTTF} = \frac{1}{\frac{100}{10^9}} \][/tex]
[tex]\[ \text{MTTF} = \frac{10^9}{100} \][/tex]
[tex]\[ \text{MTTF} = 10^7 \text{ hours} \][/tex]
Part (b): Availability
Availability (A) is defined as the ratio of the system's uptime to the total time (uptime + downtime). It's given by the formula:
[tex]\[ A = \frac{\text{MTTF}}{\text{MTTF} + \text{MTTR}} \][/tex]
Where MTTR is the Mean Time to Repair.
Given:
- MTTF = [tex]10^7[/tex] hours (from part a)
- MTTR = 1 day = 24 hours
Substitute these values into the formula:
[tex]\[ A = \frac{10^7}{10^7 + 24} \][/tex]
Now, calculate the availability:
[tex]\[ A = \frac{10^7}{10^7 + 24} \approx \frac{10^7}{10^7} \][/tex]
Since[tex]\( 10^7 \)[/tex] is much larger than 24, the approximation holds well:
[tex]\[ A \approx 1 \][/tex]
For a more precise calculation:
[tex]\[ A = \frac{10^7}{10^7 + 24} = \frac{10^7}{10,000,000 + 24} = \frac{10^7}[/tex]{10,000,024}
[tex]\[ A \approx 0.9999976 \][/tex]
Part (c): MTTF for a System with 1000 Processors
In a system with 1000 processors, assuming that the failure of any single processor results in the failure of the entire system, the MTTF of the system can be found by dividing the MTTF of a single processor by the number of processors.
Given:
- MTTF (single processor) = [tex]10^7[/tex] hours (from part a)
- Number of processors = 1000
Calculate the system MTTF:
[tex]\[ \text{MTTF}_{\text{system}} = \frac{\text{MTTF}_{\text{single}}}{\text{Number of processors}} \][/tex]
Substitute the values:
[tex]\[ \text{MTTF}_{\text{system}} = \frac{10^7}{1000} \][/tex]
[tex]\[ \text{MTTF}_{\text{system}} = 10^4 \text{ hours} \][/tex]
Summary:
- (a) MTTF for a single processor: [tex]\( 10^7 \)[/tex] hours
- (b) Availability of the system: approximately 0.9999976
- (c) MTTF for a system with 1000 processors: [tex]\( 10^4 \)[/tex] hours
Complete question;
Availability is the most important consideration for designing servers, followed closely by scalability and throughput.
a. [10] <1.7> We have a single processor with a failures in time (FIT) of 100. What is the mean time to failure (MTTF) for this system?
b. [10] <1.7> If it takes 1 day to get the system running again, what is the availability of the system?
c. [20] <1.7> Imagine that the government, to cut costs, is going to build a supercomputer out of inexpensive computers rather than expensive, reliable computers. What is the MTTF for a system with 1000 processors? Assume that if one fails, they all fail.
The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 days.
1- Assume the distribution is bell-shaped (symmetric). The percent of pregnancies last between 236 and 281 days is approximately, [ Select ] ["81.5 %", "19.5%", "68%", "99.7%", "95%"]
2- - Assume the distribution is bell-shaped (symmetric). The percent of pregnancies last between 236 and 296 days is approximately, [ Select ] ["75%", "68%", "99.7%", "95%"]
3- - Assume the distribution is not bell-shaped ( non symmetric). The percent of pregnancies last between 236 and 296 days is approximately, [ Select ] ["85%", "75%", "99.7%", "88.9%", "95%"]
Answer:
a) 81.5%
b) 95%
c) 75%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 266 days
Standard Deviation, σ = 15 days
We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(between 236 and 281 days)
[tex]P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%[/tex]
b) a) P(last between 236 and 296)
[tex]P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%[/tex]
c) If the data is not normally distributed.
Then, according to Chebyshev's theorem, at least [tex]1-\dfrac{1}{k^2}[/tex] data lies within k standard deviation of mean.
For k = 2
[tex]1-\dfrac{1}{(2)^2} = 75\%[/tex]
Atleast 75% of data lies within two standard deviation for a non normal data.
Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.
The percentage of pregnancies that last between given days using a bell-shaped symmetric distribution can be determined using the standard normal distribution table.
Explanation:1- To find the percentage of pregnancies that last between 236 and 281 days, we can use the standard normal distribution table. First, we need to standardize the values using the formula: z = (x - mean) / standard deviation. For 236 days, the z-score is (236 - 266) / 15 = -2. For 281 days, the z-score is (281 - 266) / 15 = 1. The area under the standard normal distribution curve between -2 and 1 is approximately 81.5%.
2- Following the same steps as above, for 236 days, the z-score is -2. For 296 days, the z-score is (296 - 266) / 15 = 2. The area under the standard normal distribution curve between -2 and 2 is approximately 95%.
3- If the distribution is not bell-shaped and non-symmetric, we cannot use the standard normal distribution table. Therefore, we cannot determine the percentage of pregnancies that last between 236 and 296 days without additional information.
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Lan pays a semiannual premium of $400 for automobile insurance, a monthly premium of $140 for health insurance, and an annual premium of $450 for life insurance. What is the monthly expense?
Answer:
The monthly expense is $244.16
Step-by-step explanation:
Given:
automobile insurance= $400
health insurance = $140
life insurance = $450
To find:
The monthly expense =?
Solution:
The automobile insurance is Semiannual premium . so it is paid twice a year
So for a year the total automobile insurance paid is = [tex]400 \times 2[/tex] = $800
The health insurance is monthly premium. it is paid for all 12 months.
Thus the health insure for a year is = [tex]140 \times 12[/tex] = $1680
The life insurance is annual premium. so it is paid once in a year
So for a year the life insurance paid is = $450.
The total expense for a year = 800+ 1680 + 450 = 2930
Then for one month the expense will be =[tex]\frac{2930}{12}[/tex] = $244.16
Lan's monthly expense for her insurance premiums, which include automobile, health, and life insurance, is calculated by breaking down her semiannual and annual payments into monthly amounts, and adding these to her monthly health insurance payment, resulting in a total monthly expense of $244.17.
Explanation:To calculate Lan's total monthly expense for her insurance premiums, we need to consider all the different types of insurance she pays for: automobile, health, and life insurance. The payments are made on a semiannual, monthly, and annual basis respectively. Here is a step-by-step explanation to find the total monthly expense:
Automobile insurance: Lan pays $400 semiannually. There are 6 months in a semiannual period, so the monthly expense is $400 ÷ 6 = $66.67.Health insurance: The monthly payment is given directly as $140.Life insurance: Lan pays an annual premium of $450. There are 12 months in a year, so the monthly expense is $450 ÷ 12 = $37.50.To find the total monthly expense, we sum up the monthly expenses for all three types of insurance: $66.67 (auto) + $140 (health) + $37.50 (life) = $244.17.
Therefore, Lan's monthly expense for her insurance premiums is $244.17.
A special deck of 16 card has 4 that are blue, 4 yellow, 4 green and 4 red. The four cards of each color are numbered from one to four. A single is drawn at random. Find the following probabilities.a. The probability that the card drawn is a two or a four.b. The probability that the card is a two or a four, a given that is not a one.c. The probability that the card is a two or four, given that it is either a two or a three.d The probability that the card is a two or a four, given that it is red or green.
Final answer:
The probabilities in a special deck of 16 cards for drawing a card that is a two or a four are as follows: (a) 1/2, (b) 2/3, (c) 1/2, and (d) 1/2.
Explanation:
The question involves calculating probabilities from a special deck of 16 cards with four different colors and numbers.
a. Probability of Drawing a Two or a Four
Since there are 4 twos and 4 fours in the deck, the total number of favorable outcomes is 4 + 4 = 8. The total number of cards is 16. Thus, the probability of drawing a two or a four is 8/16, which simplifies to 1/2 or 0.5.
b. Probability Given that the Card is Not a One
If we know the card is not a one, we exclude the four ones from consideration, leaving us with 12 cards. Out of these, there are still 8 cards that are either a two or a four. Therefore, the probability is 8/12, which simplifies to 2/3.
c. Probability Given that the Card is a Two or a Three
Given that the card is a two or a three, there are 4 twos and 4 threes, so 8 possible cards. Out of these, 4 are twos or fours, so the probability is 4/8, which simplifies to 1/2 or 0.5.
d. Probability Given that the Card is Red or Green
Given that the card is red or green, there are 4 red and 4 green cards, so 8 possible cards. Out of these, 2 are twos and 2 are fours, leading to 4 favorable outcomes. The probability is therefore 4/8, which simplifies to 1/2 or 0.5.
Over a long period of time, the price of a candy bar rose from $0.20 to $1.20. Over the same period, the CPI rose from 150 to 300. Adjusted for overall inflation, how much did the price of the candy bar change
Answer: 200%
Step-by-step explanation:
Given : Over a long period of time, the price of a candy bar rose from $0.20 to $1.20.
Over the same period, the CPI rose from 150 to 300. , where CPI= Consumer price index.
CPI has doubled ⇒ Overall price level doubled.
The price of candy rose by [tex]\dfrac{\$1.20}{\$0.20}=6[/tex] times.
Adjusted for overall inflation , The actual price of the candy ( today )= ($0.20 ) x ( 300) ÷ (150)
=$ 0.40
Now , The change in the price of candy bar = ( New price of candy- actual price of the candy) ÷ (actual price of the candy) x 100
= [tex]\dfrac{\$1.20-\$0.40}{\$0.40}\times100=200\%[/tex]
Hence, the change in the price of the candy = 200%
The price of the candy bar changed by 200% and this can be determined by using the given data.
Given :
The price of a candy bar rose from $0.20 to $1.20. The CPI rose from 150 to 300.The following steps can be used in order to determine the price of the candy bar change:
Step 1 - First, determine how many times the price of the candy bar rose increases.
[tex]{\rm Number \; of \; Times}=\dfrac{1.20}{0.20}[/tex]
[tex]{\rm Number \; of \; Times}=6[/tex]
Step 2 - Now, determine the actual price of the candy bar rose.
[tex]{\rm Actual\; Price}=\dfrac{0.20\times 300}{150}[/tex]
[tex]{\rm Actual \; Price } = \$ 0.40[/tex]
Step 3 - Now, determine the change in the price of the candy bar rose.
[tex]{\rm Price\; Change} = \dfrac{1.20-0.40}{0.40}\times 100[/tex]
[tex]{\rm Price\; Change} = 200\%[/tex]
The price of the candy bar changed by 200%.
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Suppose that quiz scores in a beginning statistics class have a mean of 7.47.4 with a standard deviation of 0.20.2. Using Chebyshev's Theorem, state the range in which at least 88.9%88.9% of the data will reside. Please do not round your answers.
Answer:
(6.8,8)
Step-by-step explanation:
mean=7.4
standard deviation=0.2
we have to find the range in which at least 88.9% of the data will reside
1-1/k²=0.889
1/k²=1-0.889
1/k²=0.111
k²=1/0.111=9.009
k=3.002
so, k=3
The range of values can be computed as mean±k(standard deviation).
Thus, the range in which at least 88.9% of the data will reside is
(mean-k(standard deviation), mean+ k(standard deviation))
(7.4-3(0.2),7.4+3(0.2))
(7.4-0.6,7.4+0.6)
(6.8,8)
Thus, the range in which at least 88.9% of the data will reside is (6.8,8).
A horse trots in a circle around its trainer at the end of a 22-foot-long rope. Find the area of the circle that is swept out. Round to the nearest square foot.
The horse describes a circle with radius 22.
The area for a circle with radius [tex]r[/tex] is [tex]A=\pi r^2[/tex]
So, in your case, we have
[tex]A=\pi r^2 = 22\cdot 22\cdot\pi=484\pi[/tex]
Since [tex]\pi\approx 3.14[/tex], we have
[tex]A\approx 484\cdot 3.14=1519.76[/tex]
If we have to round this to the nearest square foot, we have 1520.
3-141. The time between the arrival of e-mail messages at your computer is exponentially distributed with a mean of 2 hours. (a) What is the probability that you do not receive a message during a 2-hour period?
Answer:
The probability that there was no messages received during a 2-hour period is 0.3679.
Step-by-step explanation:
Let the random variable X = time between the arrival of e-mail messages.
The random variable [tex]X\sim Exp(\lambda)[/tex]
The probability distribution function of exponential distribution is:
[tex]f(x)=\left \{ {{\lambda e^{-\lambda x};\ x>0} \atop {0};\ otherwise} \right.[/tex]
The mean of the distribution is, Mean = 2.
The value of λ is:
[tex]\lambda=\frac{1}{Mean} =\frac{1}{2}=0.50[/tex]
Compute the probability that there was no messages received during a 2-hour period as follows:
[tex]P(X>2)=1-P(X\leq 2)\\=1-\int\limits^{2}_{0} {\lambda e^{-\lambda x}} \, dx \\=1-\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{2}_{0}\\=1-[1-e^{-\frac{x}{2} }]^{2}_{0}\\=1-[1-e^{-\frac{2}{2}}]\\=e^{-1}\\=0.3679[/tex]
Thus, the probability that there was no messages received during a 2-hour period is 0.3679.
Although the rules of probability are just basic facts about percents or proportions, we need to be able to use the language of events and their probabilities. Choose an American adult aged 20 years and over at random. Define two events:
A= the person chosen is obese
B= the person chosen is overweight, but not obese
According to the National Center for Health Statistics, P(A) = 0.38 and P(B) =0.33.
a. Select the correct description stating what the event A or B is
a. A or B is the event that the person chosen is not obese or not overweight.
b. A or B is the event that the person chosen is overweight or obese or both.
c. A or B is the event that the person chosen is overweight and obese.
d. A or B is the event that the person is overweight or obese
What is P(A or B)?
a. P(A or B)= 0.02
b. P(A or B)= 0.34
c. P(A or B)= 0.71
d. P(A or B)= 0.55
c. If C is the event that the person chosen has normal weight or less, what is P(C)?
a. P(C)= 0.29
b. P(C)= 0.66
c. P(C) = 0.68
d. P(C)= 0.45
Answer:
a) Option D is correct for this question.
That is, A or B is the event that the person is overweight or obese.
But for other questions whose sets aren't disjoint, event A or B usually means all the elements that are in either A or B or both sets.
b) Option C.
P(A or B) = 0.71
c) Option A
P(C) = 1 - P(A or B) = 0.29
Step-by-step explanation:
b) Event B specifically rules out obesity, meaning, set A and set B have no elements in common.
In a normal probability question, event A or B usually means all the elements that are in either A or B and elements that are in the two sets.
But for this question, since, it has been made clear that there are no common elements in the two sets, event A or B is event that the person is overweight or obese. Option D.
b) For disjoint sets, P(A or B) = P(A) + P(B) = 0.38 + 0.33 = 0.71. Option C.
c) P(C) is the set of all elements that are not in either A or B.
P(C) = 1 - P(A or B) = 1 - 0.71 = 0.29. Option A.
The density and the specific volume of a simple compressible system are known. What is the number of additional intensive, independent properties needed to fix the state of this system?
No additional intensive properties are needed to fix the state of a simple compressible system beyond density and specific volume.
Explanation:In a simple compressible system, the state of the system can be determined by fixing the values of two intensive, independent properties such as temperature and pressure. These two properties are typically sufficient to determine the state. Therefore, no additional intensive, independent properties are needed to fix the state of the system beyond the given density and specific volume.
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To fix the state of a system, knowing the density and specific volume as intensive properties is not enough; additional independent properties are required. The number of additional properties needed depends on the components in the system.
Density and specific volume are intensive properties of a system. To fix the state of the system, we need to know the values of additional intensive, independent properties. The number of additional intensive, independent properties required to fix the state of the system depends on the number of components present in the system.
A bacteria culture starts with 880 bacteria and grows at a rate proportional to its size. After 4 hours there will be 3520 bacteria. (a) Express the population P after t hours as a function of t . Be sure to keep at least 4 significant figures on the growth rate.
Answer:
[tex]P(t) = 880(1.4142)^{t}[/tex]
Step-by-step explanation:
The bacteria's population may be expressed by the following equation:
[tex]P(t) = P_{0}(1+r)^{t}[/tex]
In which [tex]P_{0}[/tex] is the initial population, and r is the growth rate, as a decimal.
A bacteria culture starts with 880 bacteria and grows at a rate proportional to its size. After 4 hours there will be 3520 bacteria.
This means that [tex]P_{0} = 880, P(4) = 3520[/tex].
(a) Express the population P after t hours as a function of t . Be sure to keep at least 4 significant figures on the growth rate.
We have to find the growth rate, which we do applying the value of P(4) to the equation.
[tex]P(t) = P_{0}(1+r)^{t}[/tex]
[tex]3520 = 880(1+r)^{4}[/tex]
[tex](1+r)^{4} = \frac{3520}{880}[/tex]
[tex](1+r)^{4} = 4[/tex]
Applying the fourth root to both sides of the equality.
[tex]1 + r = 1.4142[/tex]
[tex]r = 0.4142[/tex]
So the equation of P(t) is:
[tex]P(t) = 880(1.4142)^{t}[/tex]
The population P after t hours for a bacteria culture growing at a rate proportional to its size is given by the exponential growth formula P(t) = 880 * e^((ln(4)/4)t), with an initial population of 880 and a growth rate determined from the population size after 4 hours.
Explanation:The student is asking for an expression of the population P after t hours for bacteria that grow at a rate proportional to their size. Since the bacteria population growth is exponential, we use the formula P(t) = P0 * e^(rt), where P0 is the initial amount of bacteria, r is the growth rate, and t is time in hours.
We are given an initial population of 880 bacteria (P0) and after 4 hours we have 3520 bacteria. Using this information, we can find the growth rate r. The formula with the known values plugged in is 3520 = 880 * e^(4r). To solve for r, we first divide both sides by 880, getting 4 = e^(4r), and then take the natural logarithm of both sides to get ln(4) = 4r. Thus r = (ln(4))/4.
Now, the formula for P after t hours can be expressed as P(t) = 880 * e^((ln(4)/4)t). This formula will give the size of the bacteria population at any time t, in hours, assuming a constant, exponential growth rate.
1. purchase a Toyota 4runner for 25,635. promised your daughter the SUV will be hers when the car value is worth 10,000. 2. the car dealer said the SUV will depreciate in value approximately 3,000 per year. 3. write a linear equation in which y represents the total value of the car and x represents the age of the car.
Answer:buy here both
Step-by-step explanation:
Answer:
y = -3000x + 25,635
Step-by-step explanation:
well if the inital value of the car is $25,635 this means that when
x = 0 y = 25,635
(0 , 25635)
this will be our first point
Now if you tell us that in 1 year depreciate in value $3,000
this means thaw when
x = 1 y = 25,635 - 3,000
x = 1 y = 22,635
(1, 22635)
Now that we have 2 points we can have the equation
First we take the slope as follows
m = (y2 - y1) / (x2 - x1)
m = (22,635 - 25,635) / (1 - 0)
m = -3000 / 1
m = -3000
after calculating the slope we have to replace it in the following formula
(y - y1) = m (x - x1)
y - 25,635 = -3000 ( x - 0)
y - 25,635 = -3000x
y = -3000x + 25,635
Finally we replace the value of y by 10000
10,000 = -3000x + 25,635
10,000 - 25,635 = -3000x
-15,635 = -3000x
-15,635/-3000 = x
5.21167 = x years
These are the years it would take for the value to be 10,000
to know the days we simply multiply by 365
5.21167 * 365 = 1902.26 days
The Quick Change Oil Company has a number of outlets in the metropolitan Seattle area. The daily number of oil changes at the Oak Street outlet in the past 20 days are: 65 98 55 62 79 59 51 90 72 56 70 62 66 80 94 79 63 73 71 85 The data are to be organized into a frequency distribution. a. How many classes would you recommend
Answer:
5 classes.
Step-by-step explanation:
You can use the [tex]2^k[/tex] rule to determine the number of classes for a frequency distribution.
The [tex]2^k[/tex] rule says that [tex]2^k\geq n[/tex] where
[tex]k[/tex] is the number of classes
[tex]n[/tex] is the number of the data points
We know that the number of data points is [tex]n[/tex] = 20.
Next, we start searching for [tex]k[/tex] so that we can get a number 2 to the [tex]k[/tex] that is larger that the number of data points.
[tex]2^4=16\\2^5=32[/tex]
This suggests that you should use 5 classes.
Final answer:
For this data, it is recommended to use 5 classes.
Explanation:
To organize the data into a frequency distribution, you need to determine the number of classes.
The recommended number of classes can vary depending on the data.
One popular rule is to use the square root of the number of data points to determine the number of classes.
In this case, there are 20 data points, so the square root is approximately 4.47.
Since you can't have a fraction of a class, you can round up to the nearest whole number, which gives you a recommendation of 5 classes.
Ask Your Teacher Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 5 + sec(x), −π 3 ≤ x ≤ π 3 , y = 7; about y = 5
Answer:
V=15.44
Step-by-step explanation:
We have a formula
V=\int^{π/3}_{-π/3} A(x) dx ,
where A(x) calculate as cross sectional.
We have:
Inner radius: 5 + sec(x) - 5= sec(x)
Outer radius: 7 - 5=2, we get
A(x)=π 2²- π· sec²(x)
A(x)=π(4-sec²(x))
Therefore, we calculate the volume V, and we get
V=\int^{π/3}_{-π/3} A(x) dx
V=\int^{π/3}_{-π/3} π(4-sec²(x)) dx
V=[ π(4x-tan(x)]^{π/3}_{-π/3}
V=π·(8π/3-2√3)
V=15.44
We use a site geogebra.org to plot the graph.
The volume of the solid obtained by rotating the region bounded by y = 5 + sec(x), -π/3 ≤ x ≤ π/3, y = 7 about y = 5 is calculated using the disc method formula in calculus, resulting in a volume of 8π²/3 cubic units.
Explanation:The region that we need to rotate around the line y=5 is bounded by the curves y=7 and y=5+sec(x) over the interval -π/3 ≤ x ≤ π/3. The resulting solid is a disc shaped object. We can find the volume of this solid using the disc method formula in calculus:
V = π ∫ [R(x)]² dx
, where R(x) = radius function. The radius function is the absolute difference between y=7 (the upper curve) and y=5 (the line of rotation), which equals 2. Therefore, the integral becomes:
V = π ∫ from -π/3 to π/3 [2]² dx
, which simplifies to
V = 4π [x] from -π/3 to π/3
. Finally, evaluate this expression by subtracting the lower limit from the upper limit, giving
V = 4π(2π/3) = 8π²/3
cubic units.
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