Answer:
Option A)
[tex]H_{0}: \mu = 18\text{ ounces}\\H_A: \mu \neq 18\text{ ounces}[/tex]
Option A)
[tex]H_{0}: p \leq 0.6\\H_A: p > 0.6[/tex]
Option A)
[tex]H_{0}: \mu \geq 7\text{ hours}\\H_A: \mu \leq 7\text{ hours}[/tex]
Step-by-step explanation:
We have to design null and alternate hypothesis for given test:
a) Test if the mean weight of cereal in a cereal box differs from 18 ounces.
Option A)
[tex]H_{0}: \mu = 18\text{ ounces}\\H_A: \mu \neq 18\text{ ounces}[/tex]
The null hypothesis means mean cereal box weight is 18 ounces and the alternate hypothesis state that it is different than 18 ounces.
b) Test if the stock price increases on more than 60% of the trading days.
Option A)
[tex]H_{0}: p \leq 0.6\\H_A: p > 0.6[/tex]
The null hypothesis states that the proportion is less than 0.6 that is stock price is less than or equal to 60% of the trading days and alternate hypothesis states that the proportion is greater than 0.6 that is stock price increases on more than 60% of the trading days.
c) Test if Americans get an average of less than seven hours of sleep.
Option A)
[tex]H_{0}: \mu \geq 7\text{ hours}\\H_A: \mu \leq 7\text{ hours}[/tex]
The null hypothesis states that the Americans get an average of greater than or equal to 7 hours of sleep where as the alternate hypothesis states that the Americans get a sleep less than 7 hours of sleep.
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]
How long will it take $2,000 to reach $3,000 when it grows at 12 percent per year? (Do not round intermediate calculations. Round "months" to 1 decimal place.)
Answer:
4.2 years
Step-by-step explanation:
assuming simple interest (see attached graphic), the following formula applies.
A = P [ 1 + (rt) ] where,
A = final amount = $3,000
P = Principal Amount = $2,000
r = annual rate = 12% = 0.12
t = time in years
Substituting the above values into the formula gives,
3000 = 2000 [ 1 + (0.12)(t) ] (divide both sides by 2000)
3000/2000 = 1 + 0.12t
(3/2) = 1 + 0.12t (subtract 1 from both sides and rearrange)
0.12t = (3/2) - 1
0.12t = (1/2) (note 1/2 = 0.5)
0.12t = 0.5 (divide both sides by 0.12)
t = 0.5 / 0.12
t = 4.166666666667
t = 4.2 years (1 dec. pl)
Answer:
It is going to take 4.2 years for $2,000 to reach $3,000.
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex].
In this problem, we have that:
[tex]P = 2000, I = 0.12[/tex]
We want to find t when [tex]T = 3000[/tex]
So
[tex]T = E + P[/tex].
[tex]3000 = E + 2000[/tex]
[tex]E = 1000[/tex]
-----------
[tex]E = P*I*t[/tex]
[tex]1000 = 2000*0.12t[/tex]
[tex]0.12t = 0.5[/tex]
[tex]t = \frac{0.5}{0.12}[/tex]
[tex]t = 4.2[/tex]
It is going to take 4.2 years for $2,000 to reach $3,000.
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
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.
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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Define a set S recursively as follows: I. BASE: (the empty word), a, and b are in S. II. RECURSION: If s ∈ S, then a. asa ∈ S b. bsb ∈ S III. RESTRICTION: No words are in S other than those derived from I and II above.(a) Give a derivation showing that bab is in S.(b) Give a derivation showing that baab is in S.(c) Use structural induction to prove that every string in S is a palindrome. If it makes things easier, you can use the notation s to denote reversing a word (e.g., abb = bba).(d) Argue that abb is not in S
Answer:
See below
Step-by-step explanation:
a) By the base case, a∈S. By the recursive step, bsb∈S if s∈S. Then, for s=a, bab∈S.
b) By the base case, ∅∈S. By the recursive step, asa∈S if s∈S. Then, for s=∅, a∅a=aa∈S. By the recursive step, bsb∈S if s∈S. Then, for s=aa, baab∈S.
c) Structural induction: We want to prove that s is palindrome for all s∈S.
Inductive basis: If s=a,b or ∅, then s is palindrome because s has either 0 or 1 characters.
Inductive hypothesis: Suppose that r∈S is a palindrome.
Inductive step: We will prove that every element constructed from r using the recursion is also a palindrome. Because of the restriction, all elements of S are constructed in this way, except for the base case. Thus, combining this with the inductive step, we will prove that every element of S is a palindrome.
Let t be an element constructed from r by recursion. Then t=ara or t=brb. If t=ara, then t is a palindrome, because reversing the word (denote the reverse word by capitals) gives T=aRa, with R being the reverse word of r. But r is a palindrome, hence r=R and T=aRa=ara=t. Again, if t=brb, T=bRb=brb=t.
We have completed the inductive step, hence by structural induction, every element of S is palindrome.
d) By recursion and the restiction, the only elements of S of length 3 are aaa,bbb,aba,bab. abb is none of those, hence abb∉S. (note that abb is not a palindrome, so by part c), abb∉S).
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.
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.
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!
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
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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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.
An AISI 1040 cold-drawn steel tube has an OD 5 50 mm and wall thickness 6 mm. What maximum external pressure can this tube withstand if the largest principal normal stress is not to exceed 80 percent of the minimum yield strength of the material?
Answer:
82.79MPa
Step-by-step explanation:
Minimum yield strength for AISI 1040 cold drawn steel as obtained from literature, Sᵧ = 490 MPa
Given, outer radius, r₀ = 25mm = 0.025m, thickness = 6mm = 0.006m, internal radius, rᵢ = 19mm = 0.019m,
Largest allowable stress = 0.8(-490) = -392 MPa (minus sign because of compressive nature of the stress)
The tangential stress, σₜ = - ((r₀²p₀)/(r₀² - rᵢ²))(1 + (rᵢ²/r²))
But the maximum tangential stress will occur on the internal diameter of the tube, where r = rᵢ
σₜₘₐₓ = -2(r₀²p₀)/(r₀² - rᵢ²)
p₀ = - σₜₘₐₓ(r₀² - rᵢ²)/2(r₀²) = -392(0.025² - 0.019²)/2(0.025²) = 82.79 MPa.
Hope this helps!!
Final answer:
The question involves calculating the maximum external pressure a cold-drawn AISI 1040 steel tube can withstand, focusing on not exceeding 80% of the steel's minimum yield strength. The maximum external pressure that this tube can withstand without exceeding 80% of the minimum yield strength of the material is approximately [tex]\(99.84 \text{ MPa}\).[/tex]
Explanation:
To determine the maximum external pressure that the steel tube can withstand without exceeding 80% of the minimum yield strength of the material, we need to consider the maximum principal stress criterion.
The maximum principal stress [tex](\(\sigma_{\text{max}}\))[/tex] occurs at the inner surface of the tube where the wall thickness is the smallest. We can calculate this stress using the formula for hoop stress [tex](\(\sigma_h\)):[/tex]
[tex]\[\sigma_h = \frac{{p \cdot D}}{{2t}}\][/tex]
We can rearrange this equation to solve for the internal pressure (\(p\)):
[tex]\[p = \frac{{2t \cdot \sigma_{\text{max}}}}{{D}}\][/tex]
Given that the largest principal normal stress [tex](\(\sigma_{\text{max}}\))[/tex] should not exceed 80% of the minimum yield strength of the material, let's denote the minimum yield strength of the material as [tex]\(S_y\)[/tex]. Therefore, we have:
[tex]\[\sigma_{\text{max}} = 0.8 \times S_y\][/tex]
We also have the dimensions of the tube:
- Outer diameter D = 50 mm
- Wall thickness t = 6 mm
Now, we need to know the minimum yield strength [tex](\(S_y\))[/tex] of AISI 1040 cold-drawn steel. For AISI 1040 steel, the minimum yield strength is typically around 414 MPa.
Let's plug in the values:
[tex]\[p = \frac{{2 \times 6 \times 0.8 \times 414}}{{50}} = \frac{{4992}}{{50}} = 99.84 \text{ MPa}\][/tex]
So, the maximum external pressure that this tube can withstand without exceeding 80% of the minimum yield strength of the material is approximately [tex]\(99.84 \text{ MPa}\).[/tex]
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.
Suppose Z is a standard normal random variable. Find the value of z subscript italic alpha divided by 2 end subscript such that P (minus z subscript alpha divided by 2 end subscript less than Z less than space z subscript alpha divided by 2 end subscript )equals 0.95
Answer:
The result for this case would be [tex] z_{\alpha/2} =1.96[/tex]
And we can verify that:
[tex] P(-1.96 <Z< 1.96) = 0.95[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
For this case we need to satisfy the following condition:
[tex] P(-z_{\alpha/2} < Z < z_{\alpha/2}) = 0.95[/tex]
Since the normal standard distribution is symmetric and the total area below the curve is 1 since we have a probability distribution, we can rewrite this expression like this
[tex] P(-z_{\alpha/2} < Z < z_{\alpha/2}) = 1-2P(Z<-z_{\alpha/2}) = 0.95[/tex]
So we can rewrite the last expression like this:
[tex] P(Z< -z_{\alpha/2}) = \frac{1-0.95}{2}= 0.025[/tex]
And we need to see on the normal standard distribution which value accumulates 0.025 of the area on the left tail. We can use the following excel code for example:
"=NORM.INV(0.025,0,1)"
And the result for this case would be [tex] z_{\alpha/2} =1.96[/tex]
And we can verify that:
[tex] P(-1.96 <Z< 1.96) = 0.95[/tex]
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
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
Consider the following statement: There exist no integers a and b such that 18a + 6b = 1. (a) Rewrite this statement using quantifiers, propositional variables, and predicates. (b) Rewrite the negation of this statement using quantifiers, propositional variables, and predicates. (c) Using contradiction, prove the statement is false.
Answer:
See details below
Step-by-step explanation:
a) Let a,b denote variables representing integers, that is, a,b∈Z. Hence (a,b) represents a pair of integers. Let p(x,y) be the predicate "18x+6y=1". The statement "There exist no integers a and b such that 18a + 6b = 1" can be rewritten as "¬((∃(a,b))(p(a,b))", where the quantifier "∃" means "there exists".
b) The negation of this statement is ¬(¬((∃(a,b))(p(a,b))). This is equivalent to (∃(a,b))(p(a,b)).
c) Aiming for a contradiction, suppose that the statement is false, that is, there exist integers a and b such that 18a + 6b = 1. Factor 2 from this equation to obtain 2(9a+3b)=1. Since a and b are integers, k=9a+3b is an integer. Therefore 2k=1 for some integer k, that is, 1 is even, which is a contradiction. Assuming that the statement was false leads to contradictions, therefore the statement must be true, i.e, there exist no integers a and b such that 18a + 6b = 1.
A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. Each small box of paper weighs 45 pounds and each large box of paper weighs 80 pounds. There were twice as many large boxes shipped as small boxes shipped and the total weight of all boxes was 1435 pounds. Determine the number of small boxes shipped and the number of large boxes shipped.
Answer:
There were 7 small boxes and 14 large boxes shipped.
Step-by-step explanation:
This problem may be solved by a system of equations:
I am going to say that:
x is the number of small boxes used
y is the number of large boxes used
There were twice as many large boxes shipped as small boxes shipped
This means that [tex]y = 2x[/tex]
Each small box of paper weighs 45 pounds and each large box of paper weighs 80 pounds. The total weight of all boxes was 1435 pounds.
This means that [tex]45x + 80y = 1435[/tex]
So we have to solve the following system:
[tex]y = 2x[/tex]
[tex]45x + 80y = 1435[/tex]
[tex]45x + 80(2x) = 1435[/tex]
[tex]205x = 1435[/tex]
[tex]x = \frac{1435}{205}[/tex]
[tex]x = 7[/tex]
[tex]y = 2x = 2(7) = 14[/tex]
There were 7 small boxes and 14 large boxes shipped.
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).
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 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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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.
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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If you plan to keep your mileage within 12,000 to 15,000 miles per year, maintain the car very well, and only keep it for about 3 years, you should _____.
Answer: 1: lease the car, 2: 4,643.46 3: 12,160.26
Step-by-step explanation:
The term lease is defined as the practice to pay an amount of money over a certain time for the use of a product.
The given sentence is correctly filled with lease the car.
A lease is a contract between owner and buyer in which buyer can rent an asset for a certain of time with a specified money.
There are few advantage of lease.
Lease does not required down payment or very small down payment.The loan payment for a new car is lower than lease for same car.You can lease expansive car rather than buy it.At the end of the lease you can return the car without worrying to selling it.Thus, the sentence is correctly fill with lease the car.
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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.
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.
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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. 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.