Answer:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
Step-by-step explanation:
For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.
We want to know if we can apply the normal approximation, so we need to check 3 conditions:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
And we can use the empirical rule to describe the distribution of percentages.
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
The 68-95-99.7 rule can be used to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years. The appropriate conditions for using this rule are met.
Explanation:The question asks to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years using the 68-95-99.7 rule. The 68-95-99.7 rule is a statistical rule that states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
In this case, the company reported that the national college freshman-to-sophomore retention rate was 66%. Assuming that the retention rate follows a normal distribution, approximately 68% of colleges would have a retention rate within one standard deviation of 66%, approximately 95% would have a retention rate within two standard deviations, and approximately 99.7% would have a retention rate within three standard deviations.
Based on these assumptions, the appropriate conditions for using the 68-95-99.7 rule in the sampling distribution model are met.
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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A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four units, and because each has identical packaging, the selection will be random. What is the probability that (a) all four units are good, (b) exactly two units are good, and (c) at least two units are good?
(a) 0.611, (b) 0.218, (c) 0.611. Probability of all good, exactly two good, and at least two good units, respectively.
To solve these probability problems, we can use combinations. Let's calculate:
Given:
Total number of microwave ovens = 12
Number of defective units = 3
Number of good units = Total - Defective = 12 - 3 = 9
(a) Probability that all four units are good:
The probability of selecting one good unit = (Number of good units) / (Total number of units) = 9/12
For all four units to be good, we need to multiply this probability four times since each selection is independent:
P(all four are good) = (9/12)× (9/12) (9/12)× (9/12) = (9/12)^4
(b) Probability that exactly two units are good:
To find this, we need to choose 2 good units out of 4 and 2 defective units out of the remaining 12 - 4 = 8 units.
Number of ways to choose 2 good units out of 4 = C(4, 2) = 6
Number of ways to choose 2 defective units out of 8 = C(8, 2) = 28
Total number of favorable outcomes = 6 × 28
Total possible outcomes = C(12, 4)
So, the probability is:
P(exactly two units are good) = (6 ×28) / C(12, 4)
(c) Probability that at least two units are good:
This includes the cases where exactly two, three, or four units are good. We've already calculated the probability for exactly two units being good.
Probability that exactly three units are good:
Similar to the previous case, calculate the combinations for choosing 3 good units out of 4 and 1 defective unit out of the remaining 8.
Number of ways to choose 3 good units out of 4 = C(4, 3) = 4
Number of ways to choose 1 defective unit out of 8 = C(8, 1) = 8
Total number of favorable outcomes for exactly three units being good = 4 ×8
Now, the probability for at least two units being good is the sum of probabilities for exactly two, three, and four units being good:
P(at least two units are good) = P(exactly two units are good) + P(exactly three units are good) + P(all four units are good)
Sure, let's calculate each probability:
(a) Probability that all four units are good:
[tex]\[ P(\text{all four are good}) = \left(\frac{9}{12}\right)^4 = \left(\frac{3}{4}\right)^4 = \frac{81}{256} \][/tex]
(b) Probability that exactly two units are good:
[tex]\[ \text{Number of ways to choose 2 good units out of 4} = C(4, 2) = \frac{4!}{2! \times (4-2)!} = 6 \][/tex]
Number of ways to choose 2 defective units out of 8 =
[tex]C(8, 2) = \frac{8!}{2! \times (8-2)!} = 28 \][/tex]
Total possible outcomes: [tex]\( C(12, 4) = \frac{12!}{4! \times (12-4)!} = 495 \)[/tex]
[tex]\[ P(\text{exactly two units are good}) = \frac{6 \times 28}{495} = \frac{168}{495} = \frac{56}{165} \][/tex]
(c) Probability that at least two units are good:
[tex]\[ P(\text{at least two units are good})[/tex]= [tex]P(\text{exactly two units are good}) + P(\text{exactly three units are good}) + P(\text{all four units are good}) \][/tex]
Now, let's calculate the probability for exactly three units being good:
[tex]\[ \text{Number of ways to choose 3 good units out of 4} = C(4, 3) = \frac{4!}{3! \times (4-3)!} = 4 \][/tex]
Number of ways to choose 1 defective unit out of 8
[tex]= C(8, 1) = \frac{8!}{1! \times (8-1)!} = 8 \][/tex]
[tex]\[ P(\text{exactly three units are good}) = \frac{4 \times 8}{495} = \frac{32}{495} \][/tex]
Now, we can calculate [tex]\( P(\text{at least two units are good}) \):[/tex]
[tex]\[ P(\text{at least two units are good}) = \frac{56}{165} + \frac{32}{495} + \frac{81}{256} \][/tex]
[tex]\[ P(\text{at least two units are good}) = \frac{56}{165} + \frac{32}{495} + \frac{81}{256} \]\[ = \frac{4480}{12870} + \frac{1056}{12870} + \frac{3165}{12870} \]\[ = \frac{4480 + 1056 + 3165}{12870} \]\[ = \frac{8696}{12870} \][/tex]
[tex]\[ \approx 0.675 \][/tex]
So, the probabilities are:
[tex]a) Probability that all four units are good: \( \frac{81}{256} \)(b) Probability that exactly two units are good: \( \frac{56}{165} \)(c) Probability that at least two units are good: \( \frac{8696}{12870} \)[/tex]
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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Find an equation of the plane that contains (3, 4, −9) and is parallel to x + y − 5z = 1. (Use x, y, and z for coordinates.)
Answer:
x + y - 5z = 52
Step-by-step explanation:
Given data:
Point on the plane (3,4,-9)
parallel to x + y -5z = 1
Finding the normal vector
(1 , 1 , -5)(x, y , z) = 1
the normal vector is (1 , 1 , -5)
formula for finding equation the plane given the normal vector (a,b,c) and point (m,n,o) is as
a(x - m) + b(y - n) + c(z-o)=0
Substituting the data we have
1(x-3) + 1( y- 4) - 5(z+9) =0
x-3 +y -4 -5z -45 = 0
x + y - 5z = 45 +4 +3
x + y - 5z = 52
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.
In a large population of adults, the mean IQ is 112 with a standard deviation of 20. Suppose 200 adults are randomly selected for a market research campaign. The distribution of the sample mean IQ is (a)exactly Normal, mean 112, standard deviation 20. (b)approximately Normal, mean 112, standard deviation 0.1. (c)approximately Normal, mean 112, standard deviation 1.414. (d)approximately Normal, mean 112, standard deviation 20. (e)exactly Normal, mean 112, standard deviation 1.414.
Answer:
(c)approximately Normal, mean 112, standard deviation 1.414.
Step-by-step explanation:
To solve this problem, we have to understand the Central Limit Theorem
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 112, \sigma = 20, n = 200[/tex]
Using the Central Limit Theorem
The distribution of the sample mean IQ is approximately Normal.
With mean 112
With standard deviation [tex]s = \frac{20}{\sqrt{200}} = 1.414[/tex]
So the correct answer is:
(c)approximately Normal, mean 112, standard deviation 1.414.
The distribution of sample means for 200 adults is approximately normal due to the Central Limit Theorem, with a mean of 112 and a standard deviation of 1.414, making option (e) the correct choice.
Explanation:The question asks about the distribution of the sample mean IQ of 200 randomly selected adults from a large population where the mean IQ is 112 with a standard deviation of 20. According to the Central Limit Theorem, when the sample size is large, the distribution of the sample means will be approximately normal (normal by the Central Limit Theorem), even if the source population itself is not perfectly normal.
The mean of the sampling distribution of the sample mean will be the same as the mean of the population, so the mean will be 112.
However, the standard deviation of the sampling distribution (often called the standard error) is equal to the standard deviation of the population divided by the square root of the sample size. So, the standard deviation for the sample mean for 200 adults would be 20 / √200, which is about 1.414. Therefore, the correct choice is (e) exactly Normal, mean 112, standard deviation 1.414.
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.
You buy three tickets to play the "Mega Big Time Jackpot" in which the prize is $200,000. The chance any ticket wins is 1 out of 614,679 and is independent of any other ticket winning, what is the chance you lose all three times? Round to two decimal places.
a. 0.95
b. 1.00
c. 0.99d. 0.00
Answer: b. 1.00
Step-by-step explanation:
Given : Total tickets = 614,679
Winning ticket = 1
Number of tickets that not winning ticket= 614,679- 1 =614,678
Since , each ticket is independent of the other.
Number of tickets you bought = 3
Then, the probability that you lose all three times will be :
[tex](\dfrac{614678}{614679})^3=0.9999\approx1.00[/tex]
Hence, the the chance you lose all three times = 1.00
Thus , the correct answer is b. 1.00 .
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.
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 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
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 = $3650
Process
1.- Write two equations that represent this problem
f + m = 7
550f + 500m = 3650
2.- Solve this system of equations by substitution
f = 7 - m
550(7 - m) + 500m = 3650
Simplify
3850 - 550m + 500m = 3650
Solve for m
- 550m + 500m = 3650 - 3850
- 50m = -200
m = -200/-50
m = 4
3.- Calculate f
f = 7 - 4
f = 3
Say you buy an house as an investment for 350000$ (assume that you did not need a mortgage). You estimate that the house will increase in value continuously by 43750$ per year. At any time in the future you can sell the house and invest the money in a fund with a yearly interest rate of 7% compounded weekly.
If you want to maximize your return, after how many years should you sell the house? Report your answer to 1 decimal place.
y years= ?
To maximize your return, you should sell the house after approximately 4.3 years.
Explanation:To maximize your return, you should sell the house when its value plus the accumulated interest in the fund is the highest. Let's calculate the number of years for this to happen:
Value of the house after y years = $350,000 + $43,750y
Value of the fund after y years = $350,000(1 + 0.07/52)^(52y)
To find the number of years that maximizes the return, we need to find the point where the two values are equal. This can be done by solving the equation:
$350,000 + $43,750y = $350,000(1 + 0.07/52)^(52y)
Solving this equation will give us the number of years to maximize the return.
After calculating the equation, we find that after approximately 4.3 years, you should sell the house to maximize your return.
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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.
HELPPPPPPPPPPPPPPPPPP IM DESPERATE
Chandra will have 55 baseball cards
Step-by-step explanation:
Total baseball cards = 152
Let
Chandra's collection = p
Chandra has 42 baseball cards less than Simon so Simon will have 42 cards more:
Simon's Collection = p+42
We need to find Chandra's collection.
Total baseball cards will be 152 so, the equation will be:
[tex]p+(p-42)=152[/tex]
Solving and finding value of p (Chandra's card)
[tex]p+p-42=152\\2p=152+42\\2p=110\\p=110/2\\p=55[/tex]
So, Chandra will have 55 baseball cards.
Keywords: Word Problems
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Answer:
Let Chandra's collection contain x baseball cards and Simon's collection contain (x+42) baseball cards
According to the above problem, we get the equation
[tex]x + (x + 42) = 152 \\ 2x + 42 = 152 \\ 2x = 152 - 42 \\ 2x = 110 \\ x = \frac{110}{2} \\ \boxed{x = 55}[/tex]
Hence, Chandra have 55 baseball cards.
55 is the right answer.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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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]
The distribution of students’ heights in a class of 100 students is normal, with a mean height of 66 inches and a standard deviation of three. With these parameters, answer the associated question(s). Between which two heights (in inches) do the middle 60 students fall? Round to the nearest tenths place if a fraction.
Answer:
The middle 60 students fall between 63.48 inches and 68.52 inches.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 66, \sigma = 3[/tex]
Between which two heights (in inches) do the middle 60 students fall?
The normal probability distribution is symmetric. So the middle 60% fall from a pvalue of 0.50 - 0.60/2 = 0.20(lower bound) to a pvalue of 0.50 + 0.60/2 = 0.80(upper bound)
Lower bound
X when Z has a pvalue of 0.20.
So X when [tex]Z = -0.84[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X - 66}{3}[/tex]
[tex]X - 66 = -0.84*3[/tex]
[tex]X = 63.48[/tex]
Upper bound
X when Z has a pvalue of 0.80.
So X when [tex]Z = 0.84[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 66}{3}[/tex]
[tex]X - 66 = 0.84*3[/tex]
[tex]X = 68.52[/tex]
The middle 60 students fall between 63.48 inches and 68.52 inches.
Final answer:
The middle 60 students in a class with normally distributed heights, a mean of 66 inches, and a standard deviation of 3 inches, fall between heights of 63.5 and 68.5 inches.
Explanation:
The student has asked about finding the range of heights for the middle 60 students in a class of 100, where the heights are normally distributed with a mean of 66 inches and a standard deviation of 3 inches. To determine this, we use the concept of percentiles in a normal distribution. Since the middle 60 students represent the 20th to 80th percentile, we need to calculate the heights corresponding to these percentiles. The Z-value associated with the 20th percentile is approximately -0.84, and for the 80th percentile, it's approximately 0.84. Using the Z-score formula Z = (X - mean) / standard deviation, we can solve for X:
For the 20th percentile: X = -0.84 x 3 + 66 = 63.5 inches
For the 80th percentile: X = 0.84 x 3 + 66 = 68.5 inches
Therefore, the middle 60 students have heights ranging from 63.5 to 68.5 inches.
HELP!!! Ryan is driving to Phoenix. Suppose that the distance to his destination (in miles) is a linear function of his total driving time (in minutes). Ryan has 58 miles to his destination after 33 minutes of driving, and he has 46.8 miles to his destination after 47 minutes of driving. How many miles will he have to his destination after 51 minutes of driving?
Answer: he have 43.6 miles to his destination after 51 minutes of driving.
Step-by-step explanation:
let x represent his total driving time (in minutes).
Let y represent the number of miles that Ryan has to his destination.
If we plot y on the vertical axis and x on the horizontal axis, a straight line would be formed. The slope of the straight line would be
Slope, m = (58 - 46.8)/(33 - 47)
m = 11.2/- 14 = - 0.8
The equation of the straight line can be represented in the slope-intercept form, y = mx + c
Where
c = intercept
m = slope
To determine the intercept, we would substitute x = 33, y = 58 and m = - 0.8 into y = mx + c. It becomes
58 = - 0.8 × 33 + c
58 = - 26.4 + c
c = 58 + 26.4
c = 84.4
The linear function becomes
y = - 0.8x + 84.4
Therefore, after 51 minutes, the number of miles that he has to his destination is
y = - 0.8 × 51 + 84.4
y = 43.6
Solve the following
4x + 2 y = 16
3x + 3y =15
answer :- y = 9/2
and x = 1/2
Answer:
Step-by-step explanation:
The given system of equations is expressed as
4x + 2y = 16 - - - - - - - - - - - - -1
3x + 3y =15- - - - - - - - - - - - - -2
We would eliminate x by multiplying equation 1 by 3 and equation 2 by 4. It becomes
12x + 6y = 48 - - - - - - - - - - - -3
12x + 12y = 60 - - - - - - - - - - - -4
Subtracting equation 4 from equation 3, it becomes
- 6y = - 12
Dividing the left hand side and the right hand side of the equation by - 6, it becomes
- 6y/-6 = - 12/ - 6
y = 2
Substituting y = 2 into equation 1, it becomes
4x + 2 × 2 = 16
4x + 4 = 16
Subtracting 4 from the left hand side and the right hand side of the equation, it becomes
4x + 4 - 4 = 16 - 4
4x = 12
Dividing the left hand side and the right hand side of the equation by 4, it becomes
4x/4 = 12/4
x = 3
For your own safety, the bank representative says that the following characteristics of PIN numbers are prohibited:________. 1) all four digits identical, 2) sequence of consecutive ascending or descending digits, such as 0123 or 3210, 3) any sequence starting with 19 or 20 (birth years). How many sequences would NOT be permitted?
Answer:
Total = N(1) +N(2)+N(3) = 10+14+200 = 224 sequences
224 sequences would NOT be permitted.
Step-by-step explanation:
1) all four digits identical
Number of possible PIN with all four digits identical N(1) = 1×1×1×10 = 10 possible PIN
2) sequence of consecutive ascending or descending digits N(2) = 14
Only Ascending starting with 0,1,2 = 3
only descending for 7,8,9 = 3
and both for 3,4,5,6 = (2×4)
Total = 3+3+(2×4) = 14
3) any sequence starting with 19 or 20 (birth years
N(3) = 2 × 1×1×10×10 = 2×100 = 200
N(3) = 200
(Note : 100 possible ways for each of 19 or 20s)
Total = N(1) +N(2)+N(3) = 10+14+200 = 224
224 sequences would NOT be permitted.
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]
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.
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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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
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]
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).
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!
It is estimated that you will pay about $80 comma 00080,000 into the social security system (FICA) over your 4040-year work span. For simplicity, assume this is an annuity of $2 comma 0002,000 per year, starting a year from today with your 2626th birthday and continuing through your 6565th birthday (your last annual contribution is on your 6565th birthday) . a. What is the future equivalent worth of your social security savings when you retire at age 6565 if the government's interest rate is 88% per year? b. What annual withdrawal can you make if you expect to live 2525 years in retirement? Let i =88% per year.
Answer:
Part a: The Future value of the annuity after 40 years is $518113.24.
Part b: The per year withdrawal in retirement for 25 years will be $48536.19.
Step-by-step explanation:
As the numbers are appearing as a duplication taking all these values as single.
Part a
Future value is given as
[tex]FV=PMT \times [\frac{{(1+I)}^{N}-1}{I}][/tex]
Here
PMT is the annual value which is $2000 per yearI is the interest rate which is given as 8%N is 40[tex]FV=PMT \times [\frac{{(1+I)}^{N}-1}{I}]\\\\FV=2000 \times [\frac{({1+.08})^{40}-1}{.08}]\\FV=\$ 518113.03[/tex]
So the Future value of the annuity after 40 years is $518113.24.
Part b
Per year withdrawal is given as
[tex]PY=\frac{Value}{\frac{1 - \frac{1}{(1+I)^N}}{I}}[/tex]
Here
PY is the per year withdrawalValue is the total amount which is $ 518113 as calculated in part aI is the rate of interest which is 8%N is 25 years as expected life to live in retirement.So the value is given as
[tex]PY=\frac{Value}{\frac{1 - \frac{1}{(1+I)^N}}{I}}\\PY=\frac{518113}{\frac{1 - \frac{1}{(1+0.08)^{25}}}{0.08}}\\PY=\frac{518113}{10.6747}\\PY=\$ 48536.19[/tex]
So the per year withdrawal in retirement for 25 years will be $48536.
To calculate the future value of the social security annuity with an 8% interest rate over 40 years, we use the future value of an annuity formula. For withdrawals, we then use the future value as the principal to determine yearly withdrawals during a 25-year retirement at the same interest rate.
The question involves calculating the future value of an annuity and determining the equivalent yearly withdrawals during retirement. For part (a), we will use the future value of annuity formula to find the future worth of the $2,000 annual payment into the social security system given an interest rate of 8% over 40 years. Part (b) involves finding the annuity payment given a fixed principal (the future value calculated in part a), an interest rate of 8%, and a withdrawal period of 25 years (the expected span of retirement).
Calculation for part (a):
The future value FV of an annuity can be calculated using the formula FV = P * [((1 + r)^n - 1) / r], where P is the payment amount per period, r is the interest rate, and n is the total number of payments. In this scenario, P = $2,000, r = 0.08 (8%), and n = 40. Plugging the numbers into the formula will give us the future value of the social security savings.
Calculation for part (b):
For withdrawals during retirement, we start with the future value calculated in part (a) as the principal amount to be entirely withdrawn over 25 years. The annuity payment A can be found using the formula A = P * [r / (1 - (1 + r)^(-n))], where P is the future value from part (a), r is again 0.08, and n is 25. This will provide the yearly withdrawal amount that can be made during retirement.