The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1999 to match the purchasing power of $1 in 1909? In other words, how much was that in 1999? (Don't use a $ sign, use 2 decimal places.)

Answer:

962,598 different groups of 5 subjects are possible.

Step-by-step explanation:

The order is not important.

For example, Math, English, Business, Geography and History is the same group as English, Math, Business, Geography and History.

So we use the combinations formula to solve this problem.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Combinations of 5 subjects from a set of 43

[tex]C_{43, 5} = \frac{43!}{5!(43-5)!} = 962598[/tex]

962,598 different groups of 5 subjects are possible.

Answer:

705,432 ways

Step-by-step explanation:

Since no girl will be left out once both teams are selected when selecting the varsity team, the junior varsity team is automatically composed by the players not selected, the number of ways to select both teams is:

[tex]n = \frac{22!}{(22-11)!11!} \\n=705,432[/tex]

There are 705,432 ways to divide the girls into a varsity team and a junior varsity team.

Answer:

The exam can be answered with exactly 8 answers correct in 6435 ways.

Step-by-step explanation:

The order is not important.

For example, answering correctly the questions 1,2,3,4,5,6,7,8 is the same outcome as answering 2,1,3,4,5,6,7,8. So we use the combinations formula to solve this problem.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

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

Combinations of 8 questions from a set of 15. So

[tex]C_{15,8} = \frac{15!}{8!(15-8)!} = 6435[/tex]

The exam can be answered with exactly 8 answers correct in 6435 ways.

To find the number of ways to answer 15 exam questions with exactly 8 correct answers, you use the binomial coefficient formula. The calculation yields 15C8 = 6,435. Thus, there are 6,435 ways to answer the exam with 8 correct answers.

The formula for finding the number of ways to choose k items from n items is given by:

nCk = n! / [k!(n-k)!]

In this case, we need to find the number of ways to get exactly 8 correct answers out of 15 questions:

Plugging these values into the formula, we get:

15C8 = 15! / [8!(15-8)!]

Which simplifies to:

15C8 = 15! / (8! * 7!)

Calculating the factorial values:

By cancelling out the common terms, we get:

15C8 = (15 × 14 × 13 × 12 × 11 × 10 × 9) / (7 × 6 × 5 × 4 × 3 × 2 × 1) = 6435

Conclusion

There are 6,435 ways to answer the exam with exactly 8 answers correct.

Answer:

a. 68% of the workers will earn between $47300 and $69700.

b. 2.5% of workers will earn above $89000

c. Approximately 0

Step-by-step explanation:

The standard normal distribution curve in the attached graph is used to solve this question.

a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.

Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range

b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900

We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.

c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.

To answer the question, we use z-scores and a z-table to find the percentages of workers who earn within certain ranges or above certain amounts.

To answer this question, we can use the concept of the standard normal distribution. First, we convert the given earnings into z-scores by subtracting the mean and dividing by the standard deviation. With these z-scores, we can then use a z-table to find the percentage of workers who earn within a certain range or above a certain amount.

a. To find the percentage of workers who earn between $47,300 and $69,700, we need to convert these values into z-scores and find the area between these two z-scores on the z-table.

b. To find the percentage of workers who earn more than $80,900, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

c. To determine how likely it is that someone earns more than $100,000, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

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Answer:

8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.

Step-by-step explanation:

Let the number of weeks of production for team A be x and for team B be y.

In a week, facility A produces 2,000 computer monitors.

In a week ,facility A produces 3,000 computer monitors.

Total monitors to be produced = 28,000

[tex]2000x+3000y=28,000[/tex]

[tex]2x+3y=28[/tex]..[1]

In a week, facility A produces 10,000 flat panel televisions.

In a week ,facility A produces 13,000 flat panel televisions.

Total flat panel televisions to be produced = 132,000

[tex]10,000x+13,000y=132,000[/tex]

[tex]10x+13y=132[/tex]..[2]

Solving equation [1] and [2] by eliminationg method.

5 × [1] + (-1) × [2]

[tex]10x+15y=140[/tex]

[tex]-10x-13y=-132[/tex]

y = 4

[tex]x = \frac{28-3\times 4}{2}=\frac{16}{2}=8[/tex]

8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.

To produce 28,000 monitors and 132,000 televisions, it would take approximately 3.5 weeks for monitors and 3.67 weeks for televisions.

To determine the number of weeks of production required, we need to divide the total number of monitors and televisions needed by the production rate of each facility.

For monitors: Facility A produces a total of 2000 + 3000 = 5000 monitors per week, and Facility B produces 3000 monitors per week. Therefore, the total production rate for monitors is 5000 + 3000 = 8000 monitors per week.

For televisions: Facility A produces a total of 10,000 + 13,000 = 23,000 televisions per week, and Facility B produces 13,000 televisions per week. Therefore, the total production rate for televisions is 23,000 + 13,000 = 36,000 televisions per week.

To produce 28,000 monitors and 132,000 televisions, it would take:

Monitors: 28,000 / 8000 = 3.5 weeks

Televisions: 132,000 / 36,000 = 3.67 weeks

Therefore, the number of weeks required to produce the desired quantities is approximately 3.5 weeks for monitors and 3.67 weeks for televisions.

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Answer:

The amount at the end of 4 years is $2,252.79.

Step-by-step explanation:

The amount formula for the compound interest compounded quarterly is:

[tex]A=P[1+\frac{r}{4}]^{4t}[/tex]

Here,

A = Amount after t years

P = Principal amount

t = number of years

r = interest rate

Given:

P = $1,250, r = 0.15, t = 4 years.

The amount at the end of 4 years is:

[tex]A=P[1+\frac{r}{4}]^{4t}\\=1250\times[1+\frac{0.15}{4}]^{4\times4}\\=1250\times1.80223\\=2252.7875\\\approx2252.79[/tex]

Thus, the amount at the end of 4 years is $2,252.79.

Final answer:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, use the formula for compound interest:

. Substituting the given values, the result is approximately $1,776.40.

Explanation:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, you can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A is the total amount of money after the specified time period

P is the principal amount (initial deposit)

r is the annual interest rate (15% in this case)

n is the number of times interest is compounded per year (4 in this case for quarterly compounding)

t is the specified time period in years (4 in this case)

Let's substitute the given values into the formula:

A ≈ 1250 * 1.82212

A ≈ $1,1776.4

Therefore, you will have approximately $1,776.40 at the end of four years.

Answer:

50% probability that a randomly selected respondent voted for Obama.

Step-by-step explanation:

We have these following probabilities:

60% probability that an Ohio resident does not have a college degree.

If an Ohio resident does not have a college degree, a 52% probability that he voted for Obama.

40% probability that an Ohio resident has a college degree.

If an Ohio resident has a college degree, a 47% probability that he voted for Obama.

What is the probability that a randomly selected respondent voted for Obama?

This is the sum of 52% of 60%(non college degree) and 47% of 40%(college degree).

So

[tex]P = 0.52*0.6 + 0.47*0.4 = 0.5[/tex]

50% probability that a randomly selected respondent voted for Obama.

Answer:

0.1029 or 10.29%

Step-by-step explanation:

P(F) =0.3

P(S) = 1-0.3 = 0.7

If missiles continue to be launched until an unsuccessful launch occurs, the probability that exactly 4 total launches will be performed is the probability that the first three launches will be successful while the fourth will be unsucessful:

[tex]P(L=4) = P(S)*P(S)*P(S)*P(F)\\P(L=4) = 0.7*0.7*0.7*0.3\\P(L=4) = 0.1029 = 10.29\%[/tex]

The probability of 10 total launches is 0.1029 or 10.29%.

Calculate the probability of three successful launches [tex](0.7^3)[/tex] followed by one unsuccessful launch (0.3), which equals 10.29%.

The subject of the question is probability, specifically related to geometric distributions. The probability of an unsuccessful missile launch is given as 0.3. To find the probability that exactly 4 total launches will be performed before the first unsuccessful launch occurs, we need to calculate the probability of having three successful launches followed by one unsuccessful launch.

The probability of a successful launch is therefore 1 - 0.3 = 0.7. Since each launch is independent, the probability of exactly three successes followed by one failure is [tex](0.7)^3[/tex] times 0.3. This is calculated as (0.7 imes 0.7 imes 0.7) times 0.3 which equals 0.1029, or 10.29% when rounded to two decimal places.

Answer:

Jenna will have to work for 210 hours

Step-by-step explanation:

i) vacation is of six days

ii) $279 for airfare

iii) $300 for food and entertainment

iv) $65 per day for her share of the hotel, for 6 days = $65 [tex]\times[/tex] 6 = $390

v) Total expenses = $279 + $300 + $390 = $969

vi) Let x be the number of hours she will have to work to save for the vacation, where she earns $25 per hours

vii) She has $550 saved for the vacation

viii) Total savings + total earnings  = Total expenses

     therefore 550 + 25x  = 969

     therefore x = (969 -550) / 25  = 209.5 hours

   therefore Jenna will have to work for 209.5 hours, or 210 hours (rounded up to the nearest whole number)

Answer:

She must work 15 more hours to pay for her vacation

Step-by-step explanation:

279+300+5(65)

579+325=904

She has already made 550 dollars so for the next part of the problem;

904-550=354

354/25=14.16

Normally we would round down but since she needs to make the minimum amount for her trip we round up to 15 hours.

Hope that helps, this is the correct answer.

Answer: it will take Sebastian's mom 45 minutes to catch up with the bus.

Step-by-step explanation:

By the time Sebastian's mom catches up with the bus, she would have covered the same distance with the bus.

Let t represent the time it will take for Sebastian's mom to catch up with the bus.

Distance = speed × time

Sebastian's school bus averages 28 miles per hour.

Distance covered by Sebastian's school bus in t hours is

28 × t = 28t

Sebastian's mom travelled at an average speed of 42mph. Since she left home 0.25 hour after Sebastian, the time it would take her to cover the same distance is

(t - 0.25) hour. Distance covered in

(t - 0.25) hour is

42(t - 0.25)

Since the distance covered is the same, then

28t = 42(t - 0.25)

28t = 42t - 10.5

42t - 28t = 10.5

14t = 10.5

t = 10.5/14

t = 0.75

Converting to minutes, it becomes

0.75 × 60 = 45 minutes

Answer:

a) not independent

b) not mutually exclusive

Step-by-step explanation:

Given:

- A 6 sided die is rolled

- Event A is rolling an even number

- Event B is rolling a prime number

Find

- (a) Are A and B independent events?

- (b) Are A and B mutually exclusive events?

Solution:

- We will find the probability of each event:

                       set(Even number: A) = {2, 4, 6} = 3 outcomes

                       set(Prime number: B) = {2 , 3, 5} = 3 outcomes

- The probabilities are:

                        P(A) = 3/6 = 0.5

                        P(B) = 3/6 = 0.5

- For Event A and B to be independent then the following condition must match:

                        P ( A & B ) = P(A)*P(B)

                        set (A&B) = {2} = 1 outcome

                        P(A&B) = 1/6

                        1 / 6 = 0.5*0.5

                        1/6 = 0.25         ...... NOT INDEPENDENT

- For Event A and B to be mutually exclusive then the following condition must match:

                        P(A&B) = 0

                        P(A&B) = 1/6

Hence, we can say the events are NOT MUTUALLY EXCLUSIVE

No, events A and B are not independent events because the probability of both events occurring is not equal to the product of their individual probabilities. Furthermore, events A and B are not mutually exclusive events because they can both occur simultaneously if the number rolled is 2.

No, events A and B are not independent events. In order for two events to be independent, the probability of both events occurring should be equal to the product of the probabilities of each event occurring individually. However, in this case, the probability of rolling an even number (event A) is 3/6 (since there are three even numbers out of six total outcomes), and the probability of rolling a prime number (event B) is 2/6 (since there are two prime numbers out of six total outcomes). Therefore, P(A and B) is not equal to P(A) * P(B), indicating that events A and B are dependent.

Moreover, events A and B are not mutually exclusive events either. Mutually exclusive events are events that cannot occur at the same time, meaning they have no outcomes in common. In this case, event A (rolling an even number) and event B (rolling a prime number) can both occur simultaneously if the number rolled is 2, which is both even and prime. Therefore, P(A and B) is not equal to 0, indicating that events A and B are not mutually exclusive.

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Answer:

answers are shown in the file attached

Step-by-step explanation:

The detailed step are as shown in the attached file.

The solution of the all the equations are:

a) x = {10 + 7k}, where k is an integer.

b) x = {2 + 23k}, where k is an integer.

c) x = {18 + 26k}, where k is an integer.

d) x = {2 + 5k}, where k is an integer.

e) x = {5 + 6k}, where k is an integer.

f) x = {1 + 6k}, where k is an integer.

To find all integer solutions (x ∈ Z) for each equation, we need to solve them using modular arithmetic.

For each equation, we will use different approaches depending on the specific form of the equation.

Let's solve them one by one:

(a) 3x ≡ 2 (mod 7):

To find x, we'll multiply both sides of the equation by the modular inverse of 3 modulo 7, which is 5 (since 3 × 5 ≡ 1 (mod 7)):

3x × 5 ≡ 2 × 5 (mod 7)

15x ≡ 10 (mod 7)

Now, we find the smallest non-negative integer solution by dividing both sides by the greatest common divisor (GCD) of 15 and 7 (which is 1 since 15 and 7 are coprime):

x ≡ 10 (mod 7)

The solution set is x = {10, 17, 24, 31, ...}.

Since we are looking for integer solutions, we can simplify it to x = {10 + 7k}, where k is an integer.

(b) 5x + 1 ≡ 13 (mod 23):

Subtract 1 from both sides:

5x ≡ 12 (mod 23)

Next, we'll find the modular inverse of 5 modulo 23, which is 14 (since 5 * 14 ≡ 1 (mod 23)):

5x × 14 ≡ 12 × 14 (mod 23)

70x ≡ 168 (mod 23)

Now, reduce the coefficients to the smallest positive residue modulo 23:

x ≡ 2 (mod 23)

The solution set is x = {2, 25, 48, ...}, which can be simplified to x = {2 + 23k}, where k is an integer.

(c) 5x + 1 ≡ 13 (mod 26):

This equation is the same as the previous one. Follow the same steps:

5x ≡ 12 (mod 26)

Find the modular inverse of 5 modulo 26, which is 21 (since 5 × 21 ≡ 1 (mod 26)):

5x × 21 ≡ 12 × 21 (mod 26)

105x ≡ 252 (mod 26)

Reduce the coefficients to the smallest positive residue modulo 26:

x ≡ 18 (mod 26)

The solution set is x = {18, 44, 70, ...}, which can be simplified to x = {18 + 26k}, where k is an integer.

(d) 9x ≡ 3 (mod 5):

To find x, we'll multiply both sides by the modular inverse of 9 modulo 5, which is 4 (since 9 × 4 ≡ 1 (mod 5)):

9x × 4 ≡ 3 × 4 (mod 5)

36x ≡ 12 (mod 5)

Reduce the coefficients to the smallest positive residue modulo 5:

x ≡ 2 (mod 5)

The solution set is x = {2, 7, 12, ...}, which can be simplified to x = {2 + 5k}, where k is an integer.

(e) 5x ≡ 1 (mod 6):

To find x, we'll multiply both sides by the modular inverse of 5 modulo 6, which is 5 (since 5 × 5 ≡ 1 (mod 6)):

5x × 5 ≡ 1 × 5 (mod 6)

25x ≡ 5 (mod 6)

Reduce the coefficients to the smallest positive residue modulo 6:

x ≡ 5 (mod 6)

The solution set is x = {5, 11, 17, ...}, which can be simplified to x = {5 + 6k}, where k is an integer.

(f) 3x ≡ 1 (mod 6):

To find x, we'll multiply both sides by the modular inverse of 3 modulo 6, which is 3 (since 3 × 3 ≡ 1 (mod 6)):

3x × 3 ≡ 1 × 3 (mod 6)

9x ≡ 3 (mod 6)

Reduce the coefficients to the smallest positive residue modulo 6:

x ≡ 1 (mod 6)

The solution set is x = {1, 7, 13, ...}, which can be simplified to x = {1 + 6k}, where k is an integer.

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Answer: Stratified sampling

Step-by-step explanation:

Stratified sampling is a particular kind of random sampling technique.

Here , the researcher splits the entire population into distinct groups known as strata.

Then he draw a sample by taking participants from each strata and continue his work on sample.

As per given ,

Researcher = Sneha

Strata = Each floor

Since from each floor she randomly selects 2 rooms and inspects them and then takes the average of these scores.

Therefore , Sneha used the Stratified sampling technique.

Answer:

The answer to your question is 2.8 x 10¹¹

Step-by-step explanation:

Data

1 tree produces 2.6 x 10² pounds of oxygen/year

number of trees = 3.9 x 10¹¹

pounds of oxygen per day = ?

Process

1.- Divide the pounds of oxygen by 365, to get the pounds of oxygen per day.

                      2.6 x 10² / 365   = 0.712

2.- Multiply the number of trees by the pounds of oxygen per day

                     3.9 x 10¹¹  x   0.712  = 2.8 x 10¹¹ pounds of oxygen

Answer:

[tex]a=285 -0.674*9.1=278.87[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 278.87.  

[tex]a=285 +0.524*9.1=289.77[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 289.77.  

The answer would be 278.87 and 289.77

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

Let X the random variable that represent the amount of energy of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(285,9.1)[/tex]  

Where [tex]\mu=285[/tex] and [tex]\sigma=9.1[/tex]

Lowest 25%

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.75[/tex]   (a)

[tex]P(X<a)=0.25[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{a-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 -0.674*9.1=278.87[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 278.87.  

Highest 30%

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.30[/tex]   (a)

[tex]P(X<a)=0.70[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=0.524<\frac{a-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 +0.524*9.1=289.77[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 289.77.  

The lower limit for the energy amounts of the remaining computers is 279.03 kwh, and the upper limit is 290.74 kwh.

To find the upper and lower limits for the energy amounts of the remaining computers, we need to find the z-scores corresponding to the lowest 25% and highest 30% of the distribution.

Using the invNorm function in a calculator, we find that the z-score for the lowest 25% is approximately -0.674 and the z-score for the highest 30% is approximately 0.524.

To find the corresponding energy amounts, we use the formula:

Lowest Energy Amount = Mean + (Z-score * Standard Deviation)

Highest Energy Amount = Mean + (Z-score * Standard Deviation)

Substituting the z-scores and the given values, we get:

Lowest Energy Amount = 285 + (-0.674 * 9.1) = 279.03 kwh

Highest Energy Amount = 285 + (0.524 * 9.1) = 290.74 kwh

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Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

Answer:

0.84 square in.

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area.

Mathematically,

C = k * A

A1 = 0.3 sq

C1 = 2500 lb

C2 = 7000 lb

C1/A1 = C2/A2

2500/0.3 = 7000/C2

= 7000 / 8333.33

= 0.84 square in.

Answer:

Jacinta was paid $127.5 for babysitting.

Step-by-step explanation:

This problem can be solved by consecutive rules of three.

Jacinta babysat for 2 1/2 hours each Saturday for 6 weeks

How many hours did she work?

She worked for 6 Saturdays, 2.5 hours in each. So

1 Saturday - 2.5 hours

6 Saturdays - x hours

[tex]x = 6*2.5[/tex]

[tex]x = 15[/tex]

She worked 15 hours.

Paid $8.50 each hour

15*$8.50 = 127.5

Jacinta was paid $127.5 for babysitting.

Answer:

-5; 4

Step-by-step explanation:

The given linear system is:

[tex]x_1+4x_2=11\\2x_1+7x_2=18[/tex]

Multiplying the first equation by -2 and adding to the second gives:

[tex]x_1+4x_2=11\\2x_1-2x_1_+7x_2-8x_2=18 -22\\\\x_1+4x_2=11\\-x_2=-4[/tex]

Multiply the second equation by 4 and add to the first to find x1:

[tex]x_1+4x_2-4x_2=11-16\\-x_2=-4\\\\x_1=-5\\x_2=4[/tex]

The order pair for the solution of the system is -5; 4.

Answer:

The system is INCONSISTENT.

Equation of two parallel lines differ only by a constant.

Step-by-step explanation:

The given equations are:

2x + y = -11

6x + 3y = 15

From the graph we see that these lines are parallel. Any two parallel lines never meet. Hence, they do not have a solution.

Hence, the system is called an Inconsistent system.

Also, note that it can be seen easily from the equations are parallel without the help of a graph.

The equation 6x + 3y = 3(2x + y)

The terms (except for the constant term) are proportional. That means they represent parallel lines.

Hence, the answer.

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove [tex]$ \forall (a, b) \in \mathbb{R}^2 $[/tex], [tex]$ (a, b) R(a, b) $[/tex].

That is, every element in the domain is related to itself.

The given relation is [tex]$\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$[/tex]

Reflexive:

[tex]$ (a, b) \sim (a, b) $[/tex] since [tex]$ a^2 + b^2 = a^2 + b^2 $[/tex]

This is true for any pair of numbers in [tex]$ \mathbb{R}^2 $[/tex]. So, [tex]$ \sim $[/tex] is reflexive.

Symmetry:

[tex]$ \sim $[/tex] is symmetry iff whenever [tex]$ (a, b) \sim (c, d) $[/tex] then [tex]$ (c, d) \sim (a, b) $[/tex].

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

[tex]$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $[/tex]

[tex]$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $[/tex]

Hence, [tex]$ (a, b) \sim (c, d) $[/tex] since [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]

Note that [tex]$ c^2 + d^2 \nleq a^2 + b^2 $[/tex]

Hence, the given relation is not symmetric.

Transitive:

[tex]$ \sim $[/tex] is transitive iff whenever [tex]$ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $[/tex] then [tex]$ (a, b) \sim (e, f) $[/tex]

To prove transitivity let us assume [tex]$ (a, b) \sim (c, d) $[/tex] and [tex]$ (c, d) \sim (e, f) $[/tex].

We have to show [tex]$ (a, b) \sim (e, f) $[/tex]

Since [tex]$ (a, b) \sim (c, d) $[/tex] we have: [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]

Since [tex]$ (c, d) \sim (e, f) $[/tex] we have: [tex]$ c^2 + d^2 \leq e^2 + f^2 $[/tex]

Combining both the inequalities we get:

[tex]$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $[/tex]

Therefore, we get:  [tex]$ a^2 + b^2 \leq e^2 + f^2 $[/tex]

Therefore, [tex]$ \sim $[/tex] is transitive.

Hence, proved.

Answer:

The answer is D) 0.476

Step-by-step explanation:

If 1 represents HIT, and 0 represents OUT, the probability that Sammy will get a hit = the number of HITS (1s) ÷ the output from the simulation (i.e., total number of HITs and OUTs in the simulation)

Where:

the number of HITS (1s) = 20

the output from the simulation (i.e., total number of HITs and OUTs in the simulation) = 42

therefore, the probability that Sammy will get a hit = 20 ÷ 42 = 0.476

This implies that the correct answer is D) 0.476

Answer:

a) the length of the wire for the circle = [tex](\frac{60\pi }{\pi+4}) in[/tex]

b)the length of the wire for the square = [tex](\frac{240}{\pi+4}) in[/tex]

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=[tex]\frac{y}{4}[/tex]

Area of the square which is L² can now be said to be;

[tex]A_S=(\frac{y}{4})^2 = \frac{y^2}{16}[/tex]

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

[tex]r = \frac{60-y}{2\pi }[/tex]

Area of the circle which is πr² can now be;

[tex]A_C= \pi (\frac{60-y}{2\pi } )^2[/tex]

     [tex]=( \frac{60-y}{4\pi } )^2[/tex]

Total Area (A);

A = [tex]A_S+A_C[/tex]

   = [tex]\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2[/tex]

For the smallest possible area; [tex]\frac{dA}{dy}=0[/tex]

∴ [tex]\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0[/tex]

If we divide through with (2) and each entity move to the opposite side; we have:

[tex]\frac{y}{18}=\frac{(60-y)}{2\pi}[/tex]

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

[tex]y= \frac{240}{\pi+4}[/tex]

∴ since [tex]y= \frac{240}{\pi+4}[/tex], we can determine for the length of the circle ;

60-y can now be;

= [tex]60-\frac{240}{\pi+4}[/tex]

= [tex]\frac{(\pi+4)*60-240}{\pi+40}[/tex]

= [tex]\frac{60\pi+240-240}{\pi+4}[/tex]

= [tex](\frac{60\pi}{\pi+4})in[/tex]

also, the length of wire for the square  (y) ; [tex]y= (\frac{240}{\pi+4})in[/tex]

The smallest possible area (A) = [tex]\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})[/tex]

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

Answer:

a) [tex] E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324[/tex]

[tex] E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348[/tex]

[tex] Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462[/tex]

b) [tex] E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5[/tex]

And replacing the result from part a we got:

[tex] E(Y) = 28*16.324 -8.5= 448.572[/tex]

c) [tex] Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208[/tex]

d) [tex] E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925[/tex]

Step-by-step explanation:

For this case we have the following probability function given:

x 13.5 15.9 19.1

p(x) 0.17 0.57 0.26

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Part a

We can calculate the expected value with the following formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And if we replace we got:

[tex] E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324[/tex]

For the second moment we can use this definition:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And if we replace we got:

[tex] E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348[/tex]

The variance is defined:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462[/tex]

Part b

For this case we define this new random variable Y = 28 X -8.5. And we want to find the expected value, so we have this:

[tex] E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5[/tex]

And replacing the result from part a we got:

[tex] E(Y) = 28*16.324 -8.5= 448.572[/tex]

Part c

For the variance we can use the following property:

[tex]Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y)[/tex]

And using this formula we have:

[tex] Var(28X -8.5) = Var(28X)+ Var(8.5)+ 2 Cov(28X,-8.5)[/tex]

The variance for a constant is 0 so then Var(8.5)=0 and Cov(28X, -8.5) = 0 since by properties if X is a random variable and a represent a constant [tex] Cov(X,a)=0[/tex], so then we just have this:

[tex] Var(28X-8.5) = Var (28X)[/tex]

Using the following property [tex] Var(aX)= a^2 Var(X)[/tex] we have:

[tex] Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208[/tex]

Part d

For this case we define [tex] H = X -0.02 X^2[/tex]

And if we find the expected value we have this:

[tex] E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925[/tex]

To compute E(X), E(X²), and V(X), multiply storage space values by their corresponding probabilities. The expected price paid is found by substituting X into the price equation and computing E(28X - 8.5). To find the expected actual capacity, substitute X into the equation h(X) = X - 0.02X² and compute E(h(X)) by multiplying each storage space value by its corresponding probability and summing up the results.

To compute the expected value (E(X)), we multiply each storage space value by its corresponding probability and sum up the results. For E(X²), we square each storage space value, multiply it by its corresponding probability, and sum up the results. To find the variance (V(X)), we subtract the square of E(X) from E(X²). The expected price paid is found by substituting X into the price equation and computing E(28X - 8.5). The variance of the price is found by substituting X into the price equation, computing V(28X - 8.5), and rounding to the nearest whole number. To find the expected actual capacity, we substitute X into the equation h(X) = X - 0.02X² and compute E(h(X)) by multiplying each storage space value by its corresponding probability and summing up the results.

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Answer:

-3/8 = -0,375

-1/3 = -0,33

Step-by-step explanation:

N/A

Answer:when you divide two number the answer is called the quotient.

Step-by-step explanation:

-3/8

=-0.375

and

-1/3

=-0.333

Answer:

The average baggage-related revenue per passenger is $18.20

Step-by-step explanation:

$25 for the first bag

$35 for the second bag

Total passenger = 100

51 passenger have no checked baggage

32 passenger have one checked baggage. Therefore the airlines charges for the 32 passenger will be: [tex]25 * 32 = 800[/tex].

17 passenger have two checked baggage. Therefore the airline charges for the 17 passenger will be: [tex](25 *17) + (35*17) = 425+595=1020[/tex]

Average baggage related revenue per passenger is: Total Revenue / passenger

Total Revenue = 1020 + 800 = $1820

Average = 1820/100 = $18.20

Answer:

The balance after 5 years will be $2,187.91.

Step-by-step explanation:

This problem can be solved by the following formula:

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

In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.

In this problem, we have that:

[tex]P = 1800, r = 0.05, t = 4[/tex]

We want to find A

So

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

[tex]A = 1800(1+0.05)^{5}[/tex]

[tex]A = 2,187.91[/tex]

The balance after 5 years will be $2,187.91.

Answer: 2188

Step-by-step explanation:

so the person deposited 1800 and expects an annual raise in the amount of 5 percent

so the equation for 1 year is

1800(1+(5/100))=the answer

but for four years u will have to power the bract by 4

180(1+(5/100))^4= 2187.9

aka 2188

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

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 = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85[/tex]

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{177.6 - 185}{3.85}[/tex]

[tex]Z = -1.92[/tex]

[tex]Z = -1.92[/tex] has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of [tex]Z = 1.92[/tex]

[tex]Z = 1.92[/tex] has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

I'm guessing that [10] refers to the set of the first 10 positive integers.

If the largest element of a given 4-permutation is 6, then the other three elements are pulled from the set {1, 2, 3, 4, 5}. This can be done in 5!/(5 - 3)! = 60 ways. Then there are four possible positions to place the 6, giving a total of 4 * 60 = 240 permutations.

If the largest element of a permutation is *at most* 6, then the maximal element is 4, 5, or 6.

Putting everything together, the total number of permutations in which the maximal element is at most 6 is 24 + 96 + 240 = 360.

The number of 4-permutations of [10] with a maximum element of 6 is 24. The number of 4-permutations of [10] with a maximum element at most 6 is 120.

To find the number of 4-permutations of [10] with a maximum element of 6, we can consider the possibilities for the position of the maximum element in the permutation. There are 4 possible positions for the maximum element: first, second, third, or fourth. If the maximum element is in the first position, the remaining 3 elements can be any combination of the remaining 3 numbers (10, 9, and 8) which gives us 3! = 6 permutations. Similarly, if the maximum element is in the second, third, or fourth position, we will have 6 permutations for each position.

Therefore, the total number of 4-permutations of [10] with a maximum element of 6 is 4 * 6 = 24.

To find the number of 4-permutations of [10] with a maximum element at most 6, we need to consider all possible values for the maximum element. We already found that there are 24 permutations with a maximum element of 6. Now, we need to consider the possibilities where the maximum element is 5, 4, 3, 2, or 1.

If the maximum element is 5, the remaining 3 elements can be any combination of the remaining 4 numbers (10, 9, 8, and 6) which gives us 4! = 24 permutations. Similarly, if the maximum element is 4, 3, 2, or 1, we will have 24 permutations for each maximum element.

Therefore, the total number of 4-permutations of [10] with a maximum element at most 6 is 24 + 24 + 24 + 24 + 24 = 120.

Answer:

a) [tex] \bar X =117.8[/tex]

[tex] Median= \frac{117+118}{2}=117.5[/tex]

The mode on this case is the most repeated value 128 with a frequency of 3

b) [tex] Range = Max -Min = 150-88=62[/tex]

[tex] s^2 = 225.334[/tex]

[tex] s= \sqrt{225.334}= 15.011[/tex]

c) [tex] y \pm s[/tex]

[tex] Lower = 117.8 -15.011=102.809[/tex]

[tex] Upper = 117.8 +15.011=132.831[/tex]

[tex] y \pm 2s[/tex]

[tex] Lower = 117.8 -2*15.011=87.797[/tex]

[tex] Upper = 117.8 +2*15.011=147.842[/tex]

[tex] y \pm 3s[/tex]

[tex] Lower = 117.8 -3*15.011=72.787[/tex]

[tex] Upper = 117.8 +3*15.011=162.85[/tex]

d) For this case we can calculate the position where we have accumulated 70% of the data below.

50*0.7 = 35

So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.

Step-by-step explanation:

For this case we consider the following data:

128,119,95,97,124,128,142,98,108,120,113,109,124,132,97,138,133,136,120,112,146,128,103,135,114,109,100,111,131,113,124,131,133,131,88,118,116,98,112,138,100,112,111,150,117,122,97,116,92,122

Part a

For this case we can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^{50} X_i}{50}[/tex]

And after replace we got [tex] \bar X =117.8[/tex]

In order to calculate the median first we order the dataset and we got:

88  92  95  97  97  97  98  98 100 100 103 108 109 109 111 111 112 112 112 113 113  114 116 116 117 118 119 120 120 122 122 124 124 124 128 128 128 131 131 131 132 133  133 135 136 138 138 142 146 150

The median would be the average between the position 25 and 26 from the data ordered and we got:

[tex] Median= \frac{117+118}{2}=117.5[/tex]

The mode on this case is the most repeated value 128 with a frequency of 3

Part b

the range is defined as the difference between the maximun and minimum so we got:

[tex] Range = Max -Min = 150-88=62[/tex]

The sample variance can be calculated with this formula:

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

And after calculate we got: [tex] s^2 = 225.334[/tex]

And the deviation is just the square root of the variance and we got:

[tex] s= \sqrt{225.334}= 15.011[/tex]

Part c

For this case we can construct the interval with 1 , 2 and 3 deviation from the mean like this:

[tex] y \pm s[/tex]

[tex] Lower = 117.8 -15.011=102.809[/tex]

[tex] Upper = 117.8 +15.011=132.831[/tex]

[tex] y \pm 2s[/tex]

[tex] Lower = 117.8 -2*15.011=87.797[/tex]

[tex] Upper = 117.8 +2*15.011=147.842[/tex]

[tex] y \pm 3s[/tex]

[tex] Lower = 117.8 -3*15.011=72.787[/tex]

[tex] Upper = 117.8 +3*15.011=162.85[/tex]

Part d

For this case we can calculate the position where we have accumulated 70% of the data below.

50*0.7 = 35

So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.

a. The average velocity for the time period beginning at t = 2 and lasting 0.5 seconds is -36 ft/s.

b. The average velocity for the time period beginning at t = 2 and lasting 0.1 seconds is -29.6 ft/s.

c. The average velocity for the time period beginning at t = 2 and lasting 0.05 seconds is -40.8 ft/s.

d. The average velocity for the time period beginning at t = 2 and lasting 0.01 seconds is 3.84 ft/s.

e. The estimated instantaneous velocity at t = 2 is -32 ft/s.

(a) To find the average velocity over a time period, we use the formula for average velocity:

[tex]\[ \text{Average Velocity} = \frac{\text{Change in height}}{\text{Change in time}} \][/tex]

Given [tex]\( y = 36t - 16t^2 \)[/tex], we'll find the heights at [tex]\( t = 2 \)[/tex] and [tex]\( t = 2.5 \)[/tex], then calculate the difference.

At [tex]\( t = 2 \), \( y = 36(2) - 16(2)^2 = 72 - 64 = 8 \)[/tex] ft.

At [tex]\( t = 2.5 \), \( y = 36(2.5) - 16(2.5)^2 = 90 - 100 = -10 \)[/tex] ft.

Change in height = [tex]\( -10 - 8 = -18 \)[/tex] ft.

Change in time = [tex]\( 2.5 - 2 = 0.5 \)[/tex] sec.

Average velocity = [tex]\( \frac{-18}{0.5} = -36 \)[/tex] ft/s.

(b) Following the same procedure as (a), at [tex]\( t = 2.1 \), \( y = 36(2.1) - 16(2.1)^2 = 75.6 - 70.56 = 5.04 \)[/tex] ft.

Change in height = [tex]\( 5.04 - 8 = -2.96 \)[/tex] ft.

Change in time = [tex]\( 2.1 - 2 = 0.1 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.96}{0.1} = -29.6 \)[/tex] ft/s.

(c) Continuing the process, at [tex]\( t = 2.05 \), \( y = 36(2.05) - 16(2.05)^2 = 73.8 - 67.84 = 5.96 \)[/tex] ft.

Change in height = [tex]\( 5.96 - 8 = -2.04 \)[/tex] ft.

Change in time = [tex]\( 2.05 - 2 = 0.05 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.04}{0.05} = -40.8 \)[/tex] ft/s.

(d) Applying the method to [tex]\( t = 2.01 \), \( y = 36(2.01) - 16(2.01)^2 = 72.36 - 64.3216 = 8.0384 \)[/tex] ft.

Change in height = [tex]\( 8.0384 - 8 = 0.0384 \)[/tex] ft.

Change in time = [tex]\( 2.01 - 2 = 0.01 \)[/tex] sec.

Average velocity = [tex]\( \frac{0.0384}{0.01} = 3.84 \)[/tex] ft/s.

(e) To estimate the instantaneous velocity at t = 2, we find the derivative of [tex]\( y = 36t - 16t^2 \)[/tex] with respect to [tex]\( t \)[/tex], which is [tex]\( v(t) = 36 - 32t \)[/tex]. Plugging in [tex]\( t = 2 \)[/tex],

we get [tex]\( v(2) = 36 - 32(2) = -32 \)[/tex] ft/s.