A car rental company charges a one-time application fee of 40 dollars, 55 dollars per day, and 13 cents per mile for its cars. A) Write a formula for the cost, C, of renting a car as a function of the number of days, d, and the number of miles driven, m. B) If C = f(d, m), then f(5, 600) =

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:

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:

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.

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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:

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:

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]

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.

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:

[tex]P(t) = 880(1.4142)^{t}[/tex]

Step-by-step explanation:

The bacteria's population may be expressed by the following equation:

[tex]P(t) = P_{0}(1+r)^{t}[/tex]

In which [tex]P_{0}[/tex] is the initial population, and r is the growth rate, as a decimal.

A bacteria culture starts with 880 bacteria and grows at a rate proportional to its size. After 4 hours there will be 3520 bacteria.

This means that [tex]P_{0} = 880, P(4) = 3520[/tex].

(a) Express the population P after t hours as a function of t . Be sure to keep at least 4 significant figures on the growth rate.

We have to find the growth rate, which we do applying the value of P(4) to the equation.

[tex]P(t) = P_{0}(1+r)^{t}[/tex]

[tex]3520 = 880(1+r)^{4}[/tex]

[tex](1+r)^{4} = \frac{3520}{880}[/tex]

[tex](1+r)^{4} = 4[/tex]

Applying the fourth root to both sides of the equality.

[tex]1 + r = 1.4142[/tex]

[tex]r = 0.4142[/tex]

So the equation of P(t) is:

[tex]P(t) = 880(1.4142)^{t}[/tex]

The population P after t hours for a bacteria culture growing at a rate proportional to its size is given by the exponential growth formula P(t) = 880 * e^((ln(4)/4)t), with an initial population of 880 and a growth rate determined from the population size after 4 hours.

The student is asking for an expression of the population P after t hours for bacteria that grow at a rate proportional to their size. Since the bacteria population growth is exponential, we use the formula P(t) = P0 * e^(rt), where P0 is the initial amount of bacteria, r is the growth rate, and t is time in hours.

We are given an initial population of 880 bacteria (P0) and after 4 hours we have 3520 bacteria. Using this information, we can find the growth rate r. The formula with the known values plugged in is 3520 = 880 * e^(4r). To solve for r, we first divide both sides by 880, getting 4 = e^(4r), and then take the natural logarithm of both sides to get ln(4) = 4r. Thus r = (ln(4))/4.

Now, the formula for P after t hours can be expressed as P(t) = 880 * e^((ln(4)/4)t). This formula will give the size of the bacteria population at any time t, in hours, assuming a constant, exponential growth rate.

Final answer:

The probabilities in a special deck of 16 cards for drawing a card that is a two or a four are as follows: (a) 1/2, (b) 2/3, (c) 1/2, and (d) 1/2.

Explanation:

The question involves calculating probabilities from a special deck of 16 cards with four different colors and numbers.

a. Probability of Drawing a Two or a Four

Since there are 4 twos and 4 fours in the deck, the total number of favorable outcomes is 4 + 4 = 8. The total number of cards is 16. Thus, the probability of drawing a two or a four is 8/16, which simplifies to 1/2 or 0.5.

b. Probability Given that the Card is Not a One

If we know the card is not a one, we exclude the four ones from consideration, leaving us with 12 cards. Out of these, there are still 8 cards that are either a two or a four. Therefore, the probability is 8/12, which simplifies to 2/3.

c. Probability Given that the Card is a Two or a Three

Given that the card is a two or a three, there are 4 twos and 4 threes, so 8 possible cards. Out of these, 4 are twos or fours, so the probability is 4/8, which simplifies to 1/2 or 0.5.

d. Probability Given that the Card is Red or Green

Given that the card is red or green, there are 4 red and 4 green cards, so 8 possible cards. Out of these, 2 are twos and 2 are fours, leading to 4 favorable outcomes. The probability is therefore 4/8, which simplifies to 1/2 or 0.5.

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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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) 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: 56 outcomes

Step-by-step explanation:

Tossing a die 3 times and not having the order recorded, we have the following outcome.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,4] [1,4,5] [1,4,6]

[1,5,5] [1,5,6]

[1,6,6]

[2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,4] [2,4,5] [2,4,6]

[2,5,5] [2,5,6]

[2,6,6]

[3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,4] [3,4,5] [3,4,6]

[3,5,5] [3,5,6]

[3,6,6]

[4,4,4] [4,4,5] [4,4,6]

[4,5,5] [4,5,6]

[4,6,6]

[5,5,5] [5,5,6]

[5,6,6]

[6,6,6]

Hence, By direct counting, the number of possible outcome is 56 outcomes.

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: 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:

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]

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:

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:

$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: Check the attached

Step-by-step explanation:

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:

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:

(6.8,8)

Step-by-step explanation:

mean=7.4

standard deviation=0.2

we have to find the range in which at least 88.9% of the data will reside

1-1/k²=0.889

1/k²=1-0.889

1/k²=0.111

k²=1/0.111=9.009

k=3.002

so, k=3

The range of values can be computed as mean±k(standard deviation).

Thus, the range in which at least 88.9% of the data will reside is

(mean-k(standard deviation), mean+ k(standard deviation))

(7.4-3(0.2),7.4+3(0.2))

(7.4-0.6,7.4+0.6)

(6.8,8)

Thus, the range in which at least 88.9% of the data will reside is (6.8,8).

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:

0.94

Step-by-step explanation:

System will function if both components function, so,

P(system function)=P(A∩B)=?

P(A∩B)=P(A)+P(B)-P(A∪B)

We are given that P(A)=0.98, P(B)=0.95 and P(A or B)=P(A∪B)=0.99.

P(A∩B)=0.98+0.95-0.99=1.93-0.99=0.94

P(system function)=P(A∩B)=0.94.

Thus, the probability that the system functions is 0.94 or 94%.

The probability that the system functions, requiring both A and B to function, is calculated using the formula for the probability of either event occurring. By rearranging and substituting the given probabilities, we find that the probability the system functions is 0.94.

To determine the probability that the system functions, we need to find the joint probability that both A and B function, denoted as P(A AND B). Given that the probability A functions is 0.98, and B functions is 0.95, we use the given that the probability either A or B functions (which includes the case where both function) is 0.99.

We start with the formula for the probability that either A or B functions, which is:

P(A OR B) = P(A) + P(B) − P(A AND B)

.

We can rearrange this to solve for P(A AND B):

P(A AND B) = P(A) + P(B) − P(A OR B)

.

Substituting the given probabilities, we get:

P(A AND B) = 0.98 + 0.95 − 0.99 = 0.94

.

Therefore, the probability that the system functions, which requires both A and B to function, is 0.94.

Total distance traveled from the starting point = 487 mm (or) 0.487 m

Solution:

Given data:

Length of each block = 135 mm

Distance traveled towards North = three blocks

                                                      = 135 × 3

                                                      = 405

Distance traveled towards North = 405 mm

Distance traveled towards West = two blocks

                                                      = 135 × 2

                                                      = 270

Distance traveled towards West = 270 mm

Total distance traveled

               [tex]=\sqrt{\text{Distance traveled in North}^2+\text{Distance traveled in West}^2}[/tex]

               [tex]=\sqrt{405^2+270^2}[/tex]

               [tex]=\sqrt{236925}[/tex]

               [tex]=135\sqrt{13}[/tex]

               = 487 mm (approximately)

Total distance traveled from the starting point = 487 mm

Let us convert mm to m.

1 m = 1000 mm

487 mm = 487 ÷ 1000 = 0.487 m

Total distance traveled from the starting point = 0.487 m

Answer:

4.5 and 8.5 hours.

Step-by-step explanation:

The empirical 95% rule states that, in a normal distribution, 95% of the data falls within two standard deviations bellow or above the mean. If the mean is 6.5 hours and the standard deviation is 1 hour, the interval is:

[tex]6.5-2*1\leq x \leq 6.5+2*1\\4.5\leq x \leq 8.5[/tex]

The range that contains the middle 95 % of hours slept per week night by college students is 4.5 to 8.5 hours.

Using the concept of Normal Distribution, and knowing that 95% of data falls within two standard deviations, we find that hours of sleep among students range between 4.5 and 8.5 hours.

This question involves the concept of a Normal distribution in statistics, commonly used in analyzing patterns of data spread. In this scenario, we know the sleep hours are normally distributed with a mean (average) of 6.5 hours and a standard deviation of 1 hour. The middle 95% of hours slept per week night by college students means we are looking for the range of sleep hours that falls within two standard deviations of the mean on a normal distribution curve.

Because the standard deviation is 1 hour, two standard deviations would be 2 hours. Thus, we subtract and add two standard deviations from the mean to find the range. That is, 6.5 - 2 = 4.5 hours (lower bound) and 6.5 + 2 = 8.5 hours (upper bound).

So, the middle 95% of sleep hours per week night by college students falls within the range of 4.5 to 8.5 hours.

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Answer:

a) Option D is correct for this question.

That is, A or B is the event that the person is overweight or obese.

But for other questions whose sets aren't disjoint, event A or B usually means all the elements that are in either A or B or both sets.

b) Option C.

P(A or B) = 0.71

c) Option A

P(C) = 1 - P(A or B) = 0.29

Step-by-step explanation:

b) Event B specifically rules out obesity, meaning, set A and set B have no elements in common.

In a normal probability question, event A or B usually means all the elements that are in either A or B and elements that are in the two sets.

But for this question, since, it has been made clear that there are no common elements in the two sets, event A or B is event that the person is overweight or obese. Option D.

b) For disjoint sets, P(A or B) = P(A) + P(B) = 0.38 + 0.33 = 0.71. Option C.

c) P(C) is the set of all elements that are not in either A or B.

P(C) = 1 - P(A or B) = 1 - 0.71 = 0.29. Option A.

Answer:

a. Mean doesn't change.

b. Median doesn't change.

c. Mode can change.

Step-by-step explanation:

Let us assume the data set with 10 observations

{2,6,4,3,2,6,4,9,4,7}.

Arranging data set in ascending order

{2,2,3,4,4,4,6,6,7,9}

mean=2+2+3+4+4+4+6+6+7+9/10=4.7

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Mode

The most repeated value is 4. So, mode is 4 for assumed data.

Increasing highest value by 10 and decreasing lowest value by 10

{-8,2,3,4,4,4,6,6,7,19}

a.

mean=-8+2+3+4+4+4+6+6+7+19/10=4.7

Mean doesn't change

b.

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Median doesn't change

c.

Most occurring value is still 4. But mode can change if the value the highest value becomes most concurring value.

A. The mean may or may not change, depending on the original data values. If the data set has an odd number of values and the highest and lowest values are not the same, then the mean will change.

B. The median will not change because it is the middle value of the data set, and adding or subtracting a constant value to all data points does not affect the relative order of the values. The position of the median will remain the same

C. The mode may or may not change. If the mode was the highest or the lowest value in the original data set and its frequency did not change after the adjustments, the mode will remain the same.

A.However, if the data set has an even number of values or the highest and lowest values are the same, then the mean will remain unchanged. For example:

Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Mean before: (5+8+10+12+15)/5 = 10

Mean after: (25+8+10+12+(-5))/5 = 10

B. For example: Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Median before: 10 (middle value of the ordered data set)

Median after: 10 (still the middle value of the ordered data set)

C. However, if the highest or lowest value was not the mode in the original data set and its frequency becomes the highest after the adjustments, the mode will change. For example:

Original data set: 5, 8, 10, 12, 15

After increasing highest by 10 and decreasing lowest by 10: 15+10=25, 8, 10, 12, 5-10=-5

Mode before: No mode (all values are unique)

Mode after: 10 (frequency of 10 is now higher than other values)

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