The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight should the citation designation be established

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!

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 .

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

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.

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.

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.

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

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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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

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]

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.

Thus, the sentence is correctly fill with lease the car.

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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]

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            

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.

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.

[tex]\boxed{t=11.3h}[/tex]

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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No additional intensive properties are needed to fix the state of a simple compressible system beyond density and specific volume.

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.

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 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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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]

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.

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).

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.

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

[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

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.

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]

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.

(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]

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.

To maximize your return, you should sell the house after approximately 4.3 years.

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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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.

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.  

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

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.

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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.