"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

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:

a) Mean Time to failure (MTTF) = (10^7) hours

b) Availability of the system = 1

c) Mean Time to failure for 1000 processors = 10^4 hours.

Step-by-step explanation:

a) Failures in time (FIT) is traditionally reported as failure Per billion hours Of Operation.

1 billion = (10^9)

FIT = 100/(10^9) = 10^-7

MTTF = 1/FIT = 1/(10^-7) = (10^7) hours

b) Availability of the system = MTTF/(MTTF + MTTR)

MTTR = mean time to repair = 24hours

Availability of the system = (10^7)/((10^7) + 24) = 0.9999 = 1

c) FIT = 1000 (processors) × 100 (FIT per processor) = (10^5) per billion hours of operations = (10^5)/(10^9) = 10^-4

MTTF = 1/FIT = 1/(10^-4) = (10^4) hours

QED!!

(a) MTTF for a single processor:[tex]\(10^7\)[/tex] hours.  

(b)System availability: approximately 0.9999976.  

(c) MTTF for a system with 1000 processors: [tex]\(10^4\)[/tex] hours.

Let's address each part of the problem step-by-step, providing detailed explanations and calculations.

Part (a): Mean Time to Failure (MTTF)

The Mean Time to Failure (MTTF) is calculated from the Failures in Time (FIT) rate. FIT is defined as the number of failures per billion [tex](10^9)[/tex] hours of operation.

Given:

- Failures in Time (FIT) = 100

First, convert FIT to the failure rate (λ).

[tex]\[ \lambda = \frac{\text{FIT}}{10^9} \text{ failures per hour} \][/tex]

So,

[tex]\[ \lambda = \frac{100}{10^9} \text{ failures per hour} \][/tex]

The MTTF is the reciprocal of the failure rate (λ):

[tex]\[ \text{MTTF} = \frac{1}{\lambda} \][/tex]

Now, calculate MTTF:

[tex]\[ \text{MTTF} = \frac{1}{\frac{100}{10^9}} \][/tex]

[tex]\[ \text{MTTF} = \frac{10^9}{100} \][/tex]

[tex]\[ \text{MTTF} = 10^7 \text{ hours} \][/tex]

Part (b): Availability

Availability (A) is defined as the ratio of the system's uptime to the total time (uptime + downtime). It's given by the formula:

[tex]\[ A = \frac{\text{MTTF}}{\text{MTTF} + \text{MTTR}} \][/tex]

Where MTTR is the Mean Time to Repair.

Given:

- MTTF = [tex]10^7[/tex] hours (from part a)

- MTTR = 1 day = 24 hours

Substitute these values into the formula:

[tex]\[ A = \frac{10^7}{10^7 + 24} \][/tex]

Now, calculate the availability:

[tex]\[ A = \frac{10^7}{10^7 + 24} \approx \frac{10^7}{10^7} \][/tex]

Since[tex]\( 10^7 \)[/tex] is much larger than 24, the approximation holds well:

[tex]\[ A \approx 1 \][/tex]

For a more precise calculation:

[tex]\[ A = \frac{10^7}{10^7 + 24} = \frac{10^7}{10,000,000 + 24} = \frac{10^7}[/tex]{10,000,024}

[tex]\[ A \approx 0.9999976 \][/tex]

Part (c): MTTF for a System with 1000 Processors

In a system with 1000 processors, assuming that the failure of any single processor results in the failure of the entire system, the MTTF of the system can be found by dividing the MTTF of a single processor by the number of processors.

Given:

- MTTF (single processor) = [tex]10^7[/tex] hours (from part a)

- Number of processors = 1000

Calculate the system MTTF:

[tex]\[ \text{MTTF}_{\text{system}} = \frac{\text{MTTF}_{\text{single}}}{\text{Number of processors}} \][/tex]

Substitute the values:

[tex]\[ \text{MTTF}_{\text{system}} = \frac{10^7}{1000} \][/tex]

[tex]\[ \text{MTTF}_{\text{system}} = 10^4 \text{ hours} \][/tex]

Summary:

- (a) MTTF for a single processor: [tex]\( 10^7 \)[/tex] hours

- (b) Availability of the system: approximately 0.9999976

- (c) MTTF for a system with 1000 processors: [tex]\( 10^4 \)[/tex] hours

Complete question;

Availability is the most important consideration for designing servers, followed closely by scalability and throughput.

a. [10] <1.7> We have a single processor with a failures in time (FIT) of 100. What is the mean time to failure (MTTF) for this system?

b. [10] <1.7> If it takes 1 day to get the system running again, what is the availability of the system?

c. [20] <1.7> Imagine that the government, to cut costs, is going to build a supercomputer out of inexpensive computers rather than expensive, reliable computers. What is the MTTF for a system with 1000 processors? Assume that if one fails, they all fail.

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: 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%.

For more information, refer to the link given below:

https://brainly.com/question/25834626

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

Learn more about Word Problems at:

#learnwithBrainly

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.

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.

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:

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.

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.

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.

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.

https://brainly.com/question/33126850

#SPJ3

Answer:

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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

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

Answer:

a) [tex] \bar X = 52[/tex]

[tex] Median = 55[/tex]

b) [tex]Q_1= \frac{44+44}{2}=44[/tex]

[tex]Q_3= \frac{57+60}{2}=58.5[/tex]

c) [tex] Range = Max -Min = 69-36=33[/tex]

d) [tex] s^2 =100.1429[/tex]

[tex] s= \sqrt{100.143}=10.0071[/tex]

e) [tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]

[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

Step-by-step explanation:

For this case we have the following dataset:

55 56 44 43 44 56 60 62 57 45 36 38 50 69 65

A total of 15 observations

Part a

We calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^{15} X_i}{15}[/tex]

And for this case we got [tex] \bar X = 52[/tex]

For the median we ust need to order the data on increasing way like this:

36, 38,43,44,44,45,50,55,56,56,57,60,62,65,69

Since the number of observations is an odd number the median would be on the 8 position from the dataset ordered on this case:

[tex] Median = 55[/tex]

Part b

In order to calculate the Q1 we need to select the following data:

36, 38,43,44,44,45,50,55

And the Q1 would be the average between the 4 and 5 positions like this:

[tex]Q_1= \frac{44+44}{2}=44[/tex]

And for the Q3 we select these values:

55,56,56,57,60,62,65,69

And the Q3 would be the average between the 4 and 5 positions like this:

[tex]Q_3= \frac{57+60}{2}=58.5[/tex]

Part c

The Range is defined as:

[tex] Range = Max -Min = 69-36=33[/tex]

Part d

In order to calculate the sample variance we can use the following formula:

[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And if we replace we got:

[tex] s^2 =100.1429[/tex]

And the deviation is just the square root of the variance:

[tex] s= \sqrt{100.143}=10.0071[/tex]

Part e

For this case we need to find the lower and upper limits for the boxplot given by:

[tex] Lower = Q_1 -1.5 IQR = 44-1.5(58.5-44) = 22.25[/tex]

[tex] Upper = Q_1 +1.5 IQR = 44+1.5(58.5-44) = 65.75[/tex]

A possible outlier would be the value of 69 since its above the upper limti for the boxplot.

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:

The probability that there was no messages received during a 2-hour period is 0.3679.

Step-by-step explanation:

Let the random variable X = time between the arrival of e-mail messages.

The random variable [tex]X\sim Exp(\lambda)[/tex]

The probability distribution function of exponential distribution is:

[tex]f(x)=\left \{ {{\lambda e^{-\lambda x};\ x>0} \atop {0};\ otherwise} \right.[/tex]

The mean of the distribution is, Mean = 2.

The value of λ is:

[tex]\lambda=\frac{1}{Mean} =\frac{1}{2}=0.50[/tex]

Compute the probability that there was no messages received during a 2-hour period as follows:

[tex]P(X>2)=1-P(X\leq 2)\\=1-\int\limits^{2}_{0} {\lambda e^{-\lambda x}} \, dx \\=1-\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{2}_{0}\\=1-[1-e^{-\frac{x}{2} }]^{2}_{0}\\=1-[1-e^{-\frac{2}{2}}]\\=e^{-1}\\=0.3679[/tex]

Thus, the probability that there was no messages received during a 2-hour period is 0.3679.

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:

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:

$129 000/yr

Step-by-step explanation:

Weighted average is used to answer this question.

total employees are 20

10 employees make 80 000

total earning for 10 employees = 80 000 * 10 = 800 000 (multiplying)

6 employees make 150 000

total earning for 6 employees= 150 000 * 6 =900 000 (multiplying)

4 employees make 220 000

total earning for 4 employees = 220 000 * 4 = 880 000 (multiplying)

To calculate weighted average all the totals are added and then divide by total number of employees.

weighted average = (800 000 + 900 000 + 880 000)/20

weighted average = 2580000/20

Weighted average = 129 000

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:

(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:

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:

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:

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]

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:

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: 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:

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.

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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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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(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: 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:

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

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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