Answer:
P-value or Probability value is the exact percentage where test statistics lie.The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis
Step-by-step explanation:
P-value or Probability value is the exact percentage where test statistics lie.
It also tells the probability of obtaining extreme results corresponding to our level of significance keeping in state that our null hypothesis is true or correct.
The criterion for rejecting the null hypothesis using the P-value approach is that if P-value < Level of significance , then we reject our null hypothesis i.e.
Suppose P-value is 2.33% and Level of significance is 5%, then we will reject our null hypothesis as 2.33% < 5%.
On the other hand, if P-value > Level of significance , then we cannot reject or accept our null hypothesis.
A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level.
Explanation:A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. It represents the probability of observing a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. In other words, it quantifies how likely it is for the observed data to have occurred by chance if the null hypothesis is true.
The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level (usually denoted as alpha). If the p-value is smaller than alpha, we reject the null hypothesis, indicating that there is enough evidence to support the alternative hypothesis. Alternatively, if the p-value is greater than or equal to alpha, we fail to reject the null hypothesis, suggesting that there is not enough evidence to support the alternative hypothesis.
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in her district.
A) If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
B) If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
Answer:
(A) The minimum sample size required achieve the margin of error of 0.04 is 601.
(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.
Step-by-step explanation:
Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.
(A)
The margin of error, MOE = 0.04.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601[/tex]
Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.
(B)
The margin of error, MOE = 0.02.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401[/tex]
Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.
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.
The Big Falcon Rocket (BFR or Starship) from Space X can carry approximately 220,000 pounds. If they only carried $100 bills, how much money can they carry?
Answer:
$9,979,032,100 or $9.979 billion
Step-by-step explanation:
The approximate weight of a $100 bill is 1 gram.
All of the calculations bellow assume that the volume of the bills would not be an issue and only concerns weight.
1 pound is equivalent to approximately 453.59237 grams.
The weight in grams that the Big Falcon Rocket can carry is:
[tex]W= 220,000*453.59237=99,790,321.4\ g[/tex]
Since each bill weighs 1 gram, the number of bills it could carry, rounded to nearest whole bill is 99,790,321. The total amount it could carry is:
[tex]A=99,790,321*\$100=\$9,979,032,100[/tex]
Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the Poisson distribution. Provide answers to the following to 3 decimal places. What is the probability that exactly two customers arrive in a minute? Find the probability that more than three customers arrive in a two-minute period. What is the probability that at least seven customers arrive in three minutes, given that exactly two arrive in the first minute?
Answer:
a) P(2)=0.270
b) P(X>3)=0.605
c) P=0.410
Step-by-step explanation:
We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:
[tex]\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}[/tex]
a) In this case: [tex]\lambda=2.1[/tex]
[tex]P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270[/tex]
Therefore, the probability is P(2)=0.270.
b) In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]
[tex]P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605[/tex]
Therefore, the probability is P(X>3)=0.605.
c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.
In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]
[tex]P(X\geq 5)=1-P(X<5)\\\\P(X\geq 5)=1-P(X\leq 4)\\\\P(X\geq 5)=1-\sum_{x=0}^4 \frac{4.2^x\cdot e^{-4.2}}{x!}\\\\P(X\geq 5)=1-0.590\\\\P(X\geq 5)=0.410[/tex]
Therefore, the probability is P=0.410.
Suppose X indicates the number of customers that enter a grocery store within one minute.
[tex]\to X \sim \text{Poisson}(2.1)[/tex]
X's probability mass function:
[tex]\to P(X=x)=\frac{e^{-2.1} \times 2.1^{x}}{x!} ,x =0,1,2,..[/tex]
For question 1:
The likelihood of exactly two consumers arriving in a minute:
[tex]P(X=2)=\frac{e^{-2.1} \times 2.1^{2}}{2!}= 0.270016 \approx 0.270[/tex]
For question 2:
Suppose Y represents the average group of consumers who enter a grocery shop in a two-minute period.
[tex]Y \sim \text{Poisson}(2.1\times 2) \ or\ Y \sim \text{Poisson}(4.2)[/tex]
Y's probability mass function:
[tex]P(Y = y) = \frac{e^{-4.2} \times 4.2^{y}}{y!}, y =0,1,2..[/tex]
This is possible that more than three clients will arrive within a two-minute timeframe.
[tex]= P(Y > 3) \\\\= 1 - P(Y \leq 3)\\\\=1- \Sigma^{3}_{y=0} \frac{e^{-4.2} \times 4.2^y}{y!}\\\\=1-0.395403\\\\= 0.604597\approx \ 0.605[/tex]
For question 3:
Given that exactly two customers arrive in the first minute, the likelihood of at least seven clients arriving in three minutes is
[tex]= \text{P( at least 5 customers arrive in two minutes)} \\\\= P(Y \geq 5)\\\\= 1 - P(Y<5)\\\\=1-P(Y \leq 4)\\\\= 1 - 0.589827\\\\= 0.410173 \approx \ 0.410[/tex]
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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]
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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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).
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
Taylor, Moore, and Jenkins are candidates for public office. It is estimated that Moore and Jenkins have about the same probability of winning, and Taylor is believed to be twice as likely to win as either of the others. Find the probability of each candidate winning the election.
Answer:
Taylor = 50%
Moore = 25%
Jenkins = 25%
Step-by-step explanation:
Assuming there are no other candidates and that someone has to win the election, the probabilities of Taylor, Moore, and Jenkins winning the election must add up to 1 or 100%.
[tex]T+M+J = 1\\[/tex]
Since Moore and Jenkins have about the same probability of winning, and Taylor is believed to be twice as likely to win as either of the others:
[tex]M=J\\T=2J\\J+J+2J=1\\J=0.25\\M=J=0.25\\T=2*0.25=0.5[/tex]
Taylor has a probability of 50% of winning the election.
Moore has a probability of 25% of winning the election.
Jenkins has a probability of 25% of winning the election.
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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.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.
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
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.
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 .
When running a half marathon (13.1 miles), it took Mark 6 minutes to run from mile marker 1 to mile marker 2, and 20 minutes to run from mile marker 2 to mile marker 4. How long did it take Mark to run from mile marker 1 to mile marker 4? 14/3 minutes 14 3 = 4.666666666666667. What was Mark's average speed as he ran from mile marker 1 to mile marker 4? miles per minute 71 minutes after starting the race, Mark passed mile marker 9. To complete the race in 110 minutes, what must Mark's average speed be as he travels from mile marker 9 to the finish line? miles per minute
Final answer:
It took Mark 26 minutes to run from mile marker 1 to mile marker 4. His average speed was 0.5038 miles per minute for that segment. To complete the race in 110 minutes, Mark's average speed from mile marker 9 to the finish line must be 0.4415 miles per minute.
Explanation:
To find how long it took Mark to run from mile marker 1 to mile marker 4, we need to add the times it took him to run from mile marker 1 to mile marker 2 and from mile marker 2 to mile marker 4. Mark took 6 minutes to run from mile marker 1 to mile marker 2, and 20 minutes to run from mile marker 2 to mile marker 4. So the total time it took him to run from mile marker 1 to mile marker 4 is 6 minutes + 20 minutes = 26 minutes.
To find Mark's average speed as he ran from mile marker 1 to mile marker 4, we need to divide the total distance he ran by the total time it took him. The total distance from mile marker 1 to mile marker 4 is 13.1 miles. So the average speed is 13.1 miles / 26 minutes = 0.5038 miles per minute.
To complete the race in 110 minutes, Mark has 39 minutes left after passing mile marker 9. To find his average speed as he travels from mile marker 9 to the finish line, we need to divide the remaining distance by the remaining time. The remaining distance is 26.22 miles - 9 miles = 17.22 miles. So his average speed is 17.22 miles / 39 minutes = 0.4415 miles per minute.
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.
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.
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]
A lock has two buttons: a "0" button and a "1" button.To open a door you need to push the buttons according to a preset 8-bit sequence. How many sequences are there? Suppose you press an arbitrary 8-bit sequence; what is the probability that the door opens? If the first try does not succeed in opening the door, you try another number; what is the probability of success?
Answer:
Step-by-step explanation:
The problem relates to filling 8 vacant positions by either 0 or 1
each position can be filled by 2 ways so no of permutation
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
b )
Probability of opening of lock in first arbitrary attempt
= 1 / 256
c ) If first fails , there are remaining 255 permutations , so
probability of opening the lock in second arbitrary attempt
= 1 / 255 .
There are several ways you might think you could enter numbers in WebAssign, that would not be interpreted as numbers. N.B. There may be hints in RED!!!
-You cannot have commas in numbers.
-You cannot have a space in a number.
-You cannot substitute the letter O for zero or the letter l for 1.
-You cannot include the units or a dollar sign in the number.
-You can have the sign of the number, + or -.
Which of the entries below will be interpreted as numbers?
a) 1.56e-9
b) -4.99
c) 40O0
d) 1.9435
e) 1.56 e-9
f) $2.59
g) 3.25E4
h) 5,000
i) 1.23 inches
The entries that will be interpreted as numbers are a) 1.56e-9, b) -4.99, d) 1.9435, and g) 3.25E4. Entries c), e), f), h), and i) violate the restrictions mentioned.
Explanation:The entries that will be interpreted as numbers are:
a) 1.56e-9b) -4.99d) 1.9435g) 3.25E4Entries a), b), d), and g) are written in proper scientific notation and do not violate any of the restrictions mentioned. Entry c) violates the rule of substituting the letter O for zero or the letter l for 1. Entry e) has a space between the number and the exponent, violating the rule of not having a space in a number. Entry f) includes a dollar sign in the number, which is not allowed. Entry h) has a comma, violating the rule of not having commas in numbers. Entry i) includes the units 'inches', which is not allowed.
Final answer:
Entries 1.56[tex]e^{-9[/tex], -4.99, and 1.9435 are valid numbers that would be interpreted by WebAssign, while 3.25E4 is also correct as it is proper scientific notation. (Option a, b,d,g)
Explanation:
When entering numbers in WebAssign, it is important to use the correct format to ensure that the numbers are interpreted accurately. Here are which of the entries will be interpreted as numbers:
a) 1.56[tex]e^{-9[/tex]: This is a correct representation of a number in scientific notation.
b) -4.99: This is a valid negative number.
c) 40O0: This contains a letter and would not be interpreted as a number.
d) 1.9435: This is a simple decimal number.
e) 1.56 [tex]e^{-9[/tex]: This is not correctly formatted for scientific notation due to the space.
f) $2.59: This includes a dollar sign, which is not valid for a number entry.
g) 3.25E4: This is correctly written in scientific notation.
h) 5,000: Commas are not allowed in number entries.
i) 1.23 inches: This includes units, which should not be included in a number entry.
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
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]
Use the formula . Find t for r = 33.2 m/h and d = 375.16 m.
[tex]\boxed{t=11.3h}[/tex]
Explanation:In this problem we have the following data:
[tex]r:speed \\ \\ d:distance \\ \\ t:time \\ \\ \\ r=33.2m/h \\ \\ d=375.16m \\ \\ \\ So \ our \ goal \ is \ to \ find \ t[/tex]
Speed, time and distance are related with the following formula:
[tex]r=\frac{d}{t} \\ \\ \\ Solving \ for \ t: \\ \\ t=\frac{d}{r} \\ \\ t=\frac{375.16}{33.2} \\ \\ \boxed{t=11.3h}[/tex]
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The revenue of 200 companies is plotted and found to follow a bell curve. The mean is $815.425 million with a standard deviation of $22.148 million. Would it be unusual for a randomly selected company to have a revenue between $747.89 and 818.68 million?1) It is impossible for this value to occur with this distribution of data.2) The value is unusual.
3) We do not have enough information to determine if the value is unusual.
4) The value is not unusual.
5) The value is borderline unusual.
Answer:
option 4
The value is not unusual.
Step-by-step explanation:
If the probability of revenue between $747.89 and $818.68 million is low i.e. less than 5% then we can say that it would be unusual for a randomly selected company to have a revenue between $747.89 and 818.68 million.
We are given that revenue found to follow a bell curve. Also, we are given that mean=815.425 and standard deviation=22.148.
P(747.89<x<818.68)=?
P((747.89-815.425)/22.148<z<(818.68-815.425)/22.148)=P(-3.05<z<0.15)
P(747.89<x<818.68)=0.4989+.0596=0.5585
Thus, the probability for a randomly selected company to have a revenue between $747.89 and 818.68 million is not low and so the value is not unusual.
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.
The average number of customers served by The Copy Shop during a typical morning (9am to noon) is 12. One morning, The Copy Shop has to close for 15 minutes.
What is the probability that no customers will arrive during this 15 minute period?
X = number of customers
a. X ~ binomial
b. X ~ negative binomial
c. X ~ hypergeometric
d. X ~ Poisson
Answer: d, p = 0.4493
Step-by-step explanation: this question is solved using a possion probability distribution because the event is occurring at a fixed rate.
For this question of ours the fixed rate is the fact that 12 customers visiting the shop within 15 minutes.
For this question our fixed rate (u) = 12/15 = 0.8
The probability distribution for possion is given as
P(x=r) = (e^-u * u^r) / r!
At this point x = 0 ( no customers coming to the shop)
P(x=0) =( e^-0.8 * 0.8^0)/ 0!
P(x=0) = (e^-0.8 * 1)/1
P(x=0) = e^-0.8
P(x=0) = 0.4493
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]
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
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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