Angela took a general aptitude test and scored in the 91st percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score?(b) What percentage were above?

Given the specific assignment of students to labs, there is only one possible way to assign the 20 students to the three labs.

To solve this problem, we are essentially counting the number of different ways to distribute 20 students among three labs: Adams Hall, Baker Hall, and Craig Hall. Given that 6 students will be assigned to Adams Hall and 11 to Baker Hall, we already know where 17 of the 20 students will be placed. The only students left to place are the remaining 3 students, who must go to the Craig Hall lab. Therefore these students could be assigned in only one way given the constraints of the problem. Thus, only one assignment is possible.

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Answer:

If repetition is allowed then the different call letters possible are 1404.

If repetition is allowed then the different call letters possible are 1250.

Step-by-step explanation:

The call letter are of either 2 letter or 3 letters.

And for them to be identified they should start with either E or W.

Consider the call letters with 2 letter:

Possible number of call letters starting with E : E _

The second place can be filled any of the 26 letters.

Possible number of call letters starting with W : W _

The second place can be filled any of the 26 letters.

Consider the call letters with 3 letter:

Possible number of call letters starting with E : E _ _

The second place and third place can be filled with any of the 26 letters.

Possible number of call letters starting with W : W _ _

The second place and third place can be filled with any of the 26 letters.

Total number of possible letters: 26 + 26 + (26)² + (26)² = 1404

Thus, if repetition is allowed then the different call letters possible are 1404.

Consider the call letters with 2 letter:

Possible number of call letters starting with E : E _

The second place can be filled any of the 25 letters.

Possible number of call letters starting with W : W _

The second place can be filled any of the 25 letters.

Consider the call letters with 3 letter:

Possible number of call letters starting with E : E _ _

The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.

Possible number of call letters starting with W : W _ _

The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.

Total number of possible letters: 25 + 25+ (25 × 24) + (26 × 24) = 1250

Thus, if repetition is allowed then the different call letters possible are 1250.

Final answer:

The structure of a fugue includes elements such as the subject, counter-subject, expositions, and episodes. Bach used techniques like inversion, retrograde, augmentation, and diminution to develop his fugues.

Explanation:

The structure of a fugue traditionally includes a subject, which is the main theme introduced at the beginning, a counter-subject, which complements the subject, expositions where the subject is introduced in different voices, and episodes, which are free sections providing contrast. Johann Sebastian Bach used various techniques to create musical development within his fugues, some of which include:

Inversion, where the subject is mirrored vertically.

Retrograde, where the subject is played backwards.

Augmentation, where the subject is presented with longer note values, thus slower.

Diminution, where the subject is presented with shorter note values, making it faster.

These techniques helped to create complexity and interest, ensuring that the fugue evolves musically throughout the piece.

Answer:

343 in³ - B

Step-by-step explanation:

Volume of cube = l³ = 7 in x 7 in x 7 in = 343 in³

Answer:

The number of different ways to arrange the 9 cars is 362,880.

Step-by-step explanation:

There are a total of 9 cars.

These 9 cars are to divided among 3 racing groups.

The condition applied is that there should be 3 cars in each group.

Use permutation to determine the total number of arrangements of the cars.

       There are 9 cars and 3 to be allotted to group 1.

       This can happen in [tex]^9P_{3}[/tex] ways.

        That is, [tex]^9P_{3}=\frac{9!}{(9-3)!} =504[/tex] ways.

       There are remaining 6 cars and 3 to be allotted to group 2.

       This can happen in [tex]^6P_{3}[/tex] ways.

        That is, [tex]^6P_{3}=\frac{6!}{(6-3)!} =120[/tex] ways.

       There are remaining 3 cars and 3 to be allotted to group 3.

       This can happen in [tex]^3P_{3}[/tex] ways.

        That is, [tex]^3P_{3}=\frac{3!}{(3-3)!} =6[/tex] ways.

The total number of ways to arrange the 9 cars is: [tex]^9P_{3}\times ^6P_{3}\times ^3P_{3}=504\times120\times6=362880[/tex]

Thus, the number of different ways to arrange the 9 cars is 362,880.

Answer:

The optimal Hedge Ratio is 0.7305.

Step-by-step explanation:

Optimal Hedge ratio is given as

[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}[/tex]

Here

So the Hedge Ratio is given as

[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}\\HR_{optimal}=0.86 \times \frac{0.79}{0.93}\\HR_{optimal}= \$0.7305[/tex]

So the optimal Hedge Ratio is 0.7305.

Answer:

[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]

Step-by-step explanation:

The general equation of a circle is as follows:

[tex](x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}[/tex]

In which the center is [tex](x_{c}, y_{c})[/tex], and r is the radius.

In this problem, we have that:

[tex]x_{c} = 4, y_{c} = 3[/tex]

So

[tex](x - 4)^{2} + (y - 3)^{2} = r^{2}[/tex]

Passing through ​(2​,-4​)

We replace into the equation to find the radius.

[tex](2 - 4)^{2} + (-4 - 3)^{2} = r^{2}[/tex]

[tex]4 + 49 = r^{2}[/tex]

[tex]r^{2} = 53[/tex]

The equation of the circle is:

[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]

Answer:

The 7th percentile is 34.2

Step-by-step explanation:

Problems of normally distributed samples are 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 = 46, \sigma = 8[/tex]

Find the 7 th percentile.

The value of X when Z has a pvalue of 0.07. So it is X when Z = -1.475.

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

[tex]-1.475 = \frac{X - 46}{8}[/tex]

[tex]X - 46 = -1.475*8[/tex]

[tex]X = 34.2[/tex]

The 7th percentile is 34.2

Answer:

They will need 200kg of pine logs.

Gauge pressure in the tank = 24045.29 Pa

Absolute pressure= 125370.29Pa

Step-by-step explanation:

Full question:sp.gr=0.8.The fluid in the nanometer is ethyl iodide with sp.gr =1.93. The manometric fluid height is 50 inches. What is the gauge pressure and absolute pressure in the tank?

Converting 50inches to metres. Multiply by 0.0254

50×0.0254=1.27m

Gauge pressure in tank is given by:

P = pgh

Where p= density,g= acceleration due to gravity, h= height

P= 1.93× 1000× 9.81× 1.27= 24045.29Pa

Absolute pressure=Ptotal= Pliquid + Patm

Absolute pressure =24045.29 + 101325=125370.29pa

Answer:

a) P(X = 0) for raisins = 0.33287

b) P(X = 2) for chocolate chips = 0.14379

c) P(X ≥ 2) for both bits = 0.59399

Step-by-step explanation:

The average amount of raisin per cookie is 600/500 = 1.2

The average amount of chocolate chips per cookie = 400/500 = 0.8

a) Using Poisson's distribution function

P(X = x) = (e^-λ)(λˣ)/x!

For raisin, Mean = λ = 1.1

x = 0

P(X = 0) = (e⁻¹•¹)(1.1⁰)/(0!) = 0.33287

b) Using Poisson's distribution function

P(X = x) = (e^-λ)(λˣ)/x!

For chocolate chips, Mean = λ = 0.8

x = 2

P(X = 2) = (e⁻⁰•⁸)(0.8²)/(2!) = 0.14379

c) the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.

For this, the average number of bit in a raisin = (600+400)/500 = 2 bits per raisin.

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]

For P(X = 0)

P(X = x) = (e^-λ)(λˣ)/x!

Mean = λ = 2

x = 0

P(X = 0) = (e⁻²)(2⁰)/(0!) = 0.13534

For P(X = 1)

P(X = x) = (e^-λ)(λˣ)/x!

Mean = λ = 2

x = 1

P(X = 1) = (e⁻²)(2¹)/(1!) = 0.27067

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]

P(X ≥ 2) = 1 - (0.13534 + 0.27067) = 1 - 0.40601 = 0.59399

Using the Poisson distribution, it is found that there is a:

a) 0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.

b) 0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.

c) 0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

The parameters are:

Item a:

600 raisins in 500 cookies, hence, the mean is:

[tex]\mu = \frac{600}{500} = 1.2[/tex]

The probability is P(X = 0), hence:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-1.2}1.2^{0}}{(0)!} = 0.3012[/tex]

0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.

Item b:

400 chips in 500 cookies, hence, the mean is:

[tex]\mu = \frac{400}{500} = 0.8[/tex]

The probability is P(X = 2), hence:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 2) = \frac{e^{-0.8}0.8^{2}}{(2)!} = 0.1438[/tex]

0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.

Item c:

1000 bits, as 600 + 400 = 1000, in 500 cookies, hence, the mean is:

[tex]\mu = \frac{1000}{500} = 2[/tex]

The probability is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-2}2^{0}}{(0)!} = 0.1353[/tex]

[tex]P(X = 1) = \frac{e^{-2}2^{1}}{(1)!} = 0.2707[/tex]

Hence:

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.4060[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594[/tex]

0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.

A similar problem is given at https://brainly.com/question/16912674

Answer: $754.94

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

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

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 500

r = 2.75% = 2.75/100 = 0.0275

n = 12 because it was compounded monthly which means 12 times in a year.

t = 15 years

Therefore,.

A = 500(1+0.0275/12)^12 × 15

A = 500(1+0.0023)^180

A = 500(1.0023)^180

A = $754.94

Answer:

No.

Step-by-step explanation:

The equipment wll save $35,000 per over for 10 years, which totals to $350,000.

If the equipment is bought on a simple interest rate of 12% annually for ten years, it will cost:

[tex]A = P (1 + rt),\\where\ A =\ Final\ amount,\ r=rate\of\interest\annually,\ t=time,\ P= Principal\ value\\\\P = 200,000r = 0.12t = 10 \\\\A = 200,000 ( 1 + 0.12\times 10)\\A = 200,000 (1 + 1.2)\\A = 200,000 (2.2)\\A = 440,000[/tex]

We will need to pay $440,000 in total for the machine in over ten years.

If we compare both values, it can be deduced that industrial equipment is more expensive than labor cost.

The decision to purchase the equipment depends on the calculation of Net Present Value, which factors in the cost of the equipment, annual saving, and interest rate. If the calculated NPV is greater than zero, it's worth investing, otherwise, it's not.

The subject of this question falls under the category of Business Finance, specifically the concept of Net Present Value (NPV). Given the costs of the equipment, the annual saving, and the interest rate (assumed to be 12%), we can calculate NPV. The net savings account for an annuity, which we assume to be $35,000 annually, and we end up with an NPV of cells (= 35000*(1-(1+0.12)^-10)/0.12 -200000). This value helps determine whether the investment is worthwhile. If the NPV is greater than 0, you should purchase the equipment as it indicates that the project's return exceeds the cost. However, if the NPV is less than 0, you should not purchase the equipment as it indicates that the project's cost exceeds the expected return.

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the property of society getting the most from its scarce recources ... Fairness might require that everyone get an equal share because they were ... If you choose to use the $20 to go to the football game, your opportunity cost of going to the game is. $20 because you could have used that money on other things plus the value

The opportunity cost of going to the football game is $20 because you could have used that money to buy other things.

The correct answer to this question is b. $20 (because you could have used the $20 to buy other things).

Opportunity cost is the value of the next best alternative that you give up when making a decision. In this case, if you choose to use the $20 to go to the football game, you are giving up the opportunity to buy other things with that money.

There are no additional costs like the value of your time spent at the game or the cost of dinner because they are not explicitly mentioned in the question.

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Answer:

Given that an article suggests

that a Poisson process can be used to represent the occurrence of

structural loads over time. Suppose the mean time between occurrences of

loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a

4-year period? c). How long must a time period be so that the probability of no loads

occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a

4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads

occurring during that period is at most 0.3 is given by: 3.3 years

Step-by-step explanation:

Answer:

[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]

[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]

[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]

Step-by-step explanation:

You draw one card at random from a standard deck of 52 playing cards

Total cards = 52

5 even numbers in each suit. 4 times 5 = 20 even number cards

[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]

(b) the card is a heart or a diamond

there are 13 heart and 13 diamonds

[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]

(c) the card is a nine or a face card

There are 4 nine and 12 face cards

[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]

Answer:

(a) 0.0492

(b) 0.0984

Step-by-step explanation:

The probability distribution of blood type in the US is:

[tex]\begin{array}{cc}Type&Probability\\O&0.46\\A&0.41\\B&0.12\\AB&0.01\end{array}[/tex]

(a) The probability that the wife has type A and the husband has type B is:

[tex]P(w=A\ and\ h=B) =0.41*0.12\\P(w=A\ and\ h=B) =0.0492[/tex]

(b) The probability that one of the couple has type A blood and the other has type B is given by the probability that the wife has type A and the husband has type B added to the probability that the wife has type B and the husband has type A.

[tex]P= P(w=A\ and\ h=B) + P(w=B\ and\ h=A)\\P=0.41*0.12+0.12*0.41=0.0984[/tex]

Final answer:

The probability that the wife has type A blood and the husband has type B blood is 4.92%. The probability that one of the couple has type A blood and the other has type B blood is 9.84%.

Explanation:

The student is being asked to calculate probabilities related to the blood types of a randomly chosen married couple in the United States using the given distribution of ABO blood types. The probabilities for each blood type are: type O at 0.46, type A at 0.41, type B at 0.12, and type AB at 0.01. To solve these questions, we use the principle that the probability of independent events occurring together is the product of their respective probabilities.

Answer:

1) Null hypothesis:[tex]p=0.53[/tex]  

2)Alternative hypothesis:[tex]p \neq 0.53[/tex]  

3) [tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]  

4) We assume that [tex]\alpha=0.05[/tex]

[tex]p_v =2*P(z<-2.605)=0.0092[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53.  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

[tex]\hat p=0.4[/tex] estimated proportion of people who would vote for the incumbent

[tex]p_o=0.53[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level  (assumed)

Confidence=95% or 0.95  (Assumed)

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.53 or not.:  

1) Null hypothesis:[tex]p=0.53[/tex]  

2)Alternative hypothesis:[tex]p \neq 0.53[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z<-2.605)=0.0092[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53 .  

Final answer:

The question involves calculating how much salt remains in a tank after 250 minutes when saltwater of a certain concentration is added and the solution is simultaneously drained at a different rate. A basic calculation suggests an addition of 1500 pounds to the initial 100 pounds, but the real answer must consider dilution and the tank's capacity, which complicates the calculation.

Explanation:

The question involves calculating the amount of salt in a tank after 250 minutes when saltwater is being pumped in and out at different rates. Initially, the tank has 100 pounds of salt and is then filled with a solution at a rate of 3 gallons per minute, with a concentration of 2 pounds of salt per gallon. Meanwhile, the tank is emptied at a rate of 1 gallon per minute.

To find the amount of salt in the tank after 250 minutes, we need to consider both the inflow and outflow of the tank. The net flow rate is 2 gallons per minute (3 gallons in - 1 gallon out). This leads to an increase in the volume of the tank by 2 gallons for every minute.

Every minute, 6 pounds of salt (3 gallons x 2 pounds/gallon) are added to the tank. However, since the tank's volume increases by 2 gallons per minute but 1 gallon is removed, we should adjust for the concentrate reduction due to dilution. Over 250 minutes, the total added salt is 250 minutes x 6 pounds/minute = 1500 pounds. However, the calculation must be adjusted to consider the dilution effect which is beyond the scope of this immediate calculation.

Initially, the tank contains 100 pounds of salt. Without accounting for dilution or maximum capacity effects, a simple addition suggests that after 250 minutes, 1600 pounds of salt would be in the tank. Yet, this simplistic approach does not account for dilution effect properly or the fact the tank has a maximum capacity, indicating that a more complex calculation is required for precise results.

Answer:

Amount paid annually = $11241.42

Step-by-step explanation:

The steps are as shown in the attached file

Answer:

where is the age three she is incorrect

Step-by-step explanation:

Answer:

she is correct its 8 people who are 3 years old and it says it

Step-by-step explanation:

Answer:

a) [tex]S(s) = 14s^2[/tex]

b) [tex]V(s) = 3s^3[/tex]

c) [tex]S(s) = \dfrac{14V(s)}{3s}[/tex]

d) 56 square meter                                    

Step-by-step explanation:

We are given the following in the question:

A closed box with a square bottom is three times high as it is wide.

Let s be the side of square and h be the height.

[tex]h = 3s[/tex]

a) Surface area of box

[tex]2(lb + bh + hl)[/tex]

where l is the length, b is the breadth and h is the height.

Putting values:

[tex]S = 2(s^2 + sh +sh)\\S = 2(s^2 + 3s^2 + 3s^2)\\S(s) = 14s^2[/tex]

b) Volume of box

[tex]l\times b \times h[/tex]

where l is the length, b is the breadth and h is the height.

Putting values:

[tex]V = s\times s\times h\\V= s\times s\times 3s\\V(s) = 3s^3[/tex]

c) Surface area in terms of volume

[tex]S(s) = 14s^2 = \dfrac{14V(s)}{3s}[/tex]

d) Surface area

Volume = 24 m³

[tex]V(s) = 24\\3s^3 = 24\\s^3 = 3\\s = 2[/tex]

[tex]S(2) = 14(2)^2 = 56\text{ square meter}[/tex]

Answer:

56 drinks

Step-by-step explanation:

Given

n = 15-4 = 11 rounds (with one additional drink)

a₀ = 3*4 = 12 drinks (between the first and the fourth round)

r = 4 (number of drinks per round)

We can use the formula

aₙ = a₀ + n*r

⇒   a₁₁ = 12 + 11*4 = 12 + 44

⇒   a₁₁ = 56 drinks

Answer: the polynomial representing the shaded region is

2x² - 6x - 1

Step-by-step explanation:

The area of the square is expressed as 4x² - 2x - 6

The area of the triangle is expressed as 2x² + 4x - 5. The area of the shaded region would be the area of the square - the area of the area of the triangle.

The polynomial that represents the area of the shaded region would be

4x² - 2x - 6 - (2x² + 4x - 5)

Collecting like terms, it becomes

= 4x² - 2x² - 2x - 4x - 6 + 5

= 2x² - 6x - 1

Final answer:

The area of the shaded region is represented by the polynomial [tex]2x^2 - 6x - 1[/tex] square inches, which is the difference between the area of the square and the area of the triangle.

Explanation:

The student asked for the polynomial that represents the area of the shaded region given the area of the square and the area of the triangle within it. To find this area, we must subtract the area of the triangle from the area of the square. The given area of the square is [tex]4x^2 - 2x - 6[/tex] square inches, and the area of the triangle is [tex]2x^2 + 4x - 5[/tex] square inches. Therefore, the polynomial representing the area of the shaded region is found by the following subtraction:

Area of the square - Area of the triangle = [tex](4x^2 - 2x - 6) - (2x^2 + 4x - 5) = 2x^2 - 6x - 1[/tex]

This simplifies to 2x^2 (from the square's area) minus 6x (the change in linear dimensions caused by the triangle) minus 1 (the constant term from completing the square). Thus, the polynomial representing the shaded region is [tex]2x^2 - 6x - 1[/tex] square inches.

Answer:

(a) P(all of the next three vehicles inspected pass) = 0.512 .

(b) P(at least one of the next three inspected fails) = 0.488 .

(c) P(exactly one of the next three inspected passes) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

Step-by-step explanation:

We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.

So, Probability that the next vehicle examined fails the inspection is 20%.

Also, it is given that successive vehicles pass or fail independently of one another.

(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection

              = 0.8 * 0.8 * 0.8 = 0.512

(b) P(at least one of the next three inspected fails) =

        1 - P(none of the next three inspected fails) = 1 - P(all next three passes)

    = 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .

(c) P(exactly one of the next three inspected passes) is given by ;

Hence, P(exactly one of the next three inspected passes) = Add all above cases ;

             (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = P(that none

     of the next three vehicle passes) + P(Only one of the next three passes)

We have calculated the P(Only one of the next three passes) in the above part of this question;

 Hence, P(at most one of the next three vehicles inspected passes) =

              (0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)

             = 0.008 + 0.096 = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;

P(All three passes/At least one of the next three vehicles passes inspection)

= P( All next three passes [tex]\bigcap[/tex] At least one of next three passes) /

      P(At least one of next three passes)

=  P( All next three passes ) / P(At least one of next three passes)

= P( All next three passes ) / 1 - P(none of the next three passes)

 =  [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .

Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

a) P(all of the next three vehicles pass) = 0.512

(b) P(at least one of the next three inspected fails) = 0.488

(c) P(exactly one of the next three inspected passes) = 0.096

(d) P(at most one of the next three vehicles inspected passes) = 0.104

(e) Given that at least one of the next three vehicles passes inspection, the probability that all three pass is approximately 1.049 (rounded to three decimal places).

Let's calculate the probabilities step by step:

(a) To find the probability that all of the next three vehicles pass, we can use the probability of a single vehicle passing (0.80) raised to the power of 3 (because the events are independent).

P(all pass) = (0.80)^3 = 0.512

(b) To find the probability that at least one of the next three vehicles fails, we can use the complement rule. It's easier to find the probability that none of them fail and then subtract that from 1.

P(at least one fails) = 1 - P(all pass) = 1 - 0.512 = 0.488

(c) To find the probability that exactly one of the next three vehicles passes, we need to consider three cases: Pass-Fail-Fail, Fail-Pass-Fail, and Fail-Fail-Pass. Each of these cases has a probability of (0.80) * (0.20) * (0.20).

P(exactly one passes) = 3 * (0.80) * (0.20) * (0.20) = 0.096

(d) To find the probability that at most one of the next three vehicles passes, we sum the probabilities of exactly one passing and none passing.

P(at most one passes) = P(none pass) + P(exactly one passes) = (0.20)^3 + 0.096 = 0.104

(e) To calculate the conditional probability that all three pass given that at least one passes, we use the formula for conditional probability:

P(all three pass | at least one passes) = P(all three pass and at least one passes) / P(at least one passes)

P(all three pass and at least one passes) = P(all pass) = 0.512 (from part a)

So, P(all three pass | at least one passes) = 0.512 / 0.488 ≈ 1.049 (rounded to three decimal places).

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Answer: G and F are mutually exclusive because they cannot occur together

Step-by-step explanation:

According to the definition of mutually exclusive events,

The events which can not occur together and probability of them occurring together is 0 are known as mutually exclusive events.

The first statement gives an implication that if one happens then other happens meaning they could both still happen so it is not true.

The second statement contradict the question about being mutually exclusive events.

The third statement also is a implication that if one event occurs then other does or does not occur.

The last statement is correct one that conforms with the question and obeys the definition of mutually exclusive events.

Answer:

  No

Step-by-step explanation:

The second and third figures are translations of each other. The first figure appears to be a reflection of the second, not a translation. Hence the first and third figures are not translations of each other.

__

If an image is translated, all of the pre-image line segments are parallel to the image line segments. That is not the case for the first and third figures, in which the top line segments go in different directions.

Answer:

b. No; it is a reflection followed by a translation.

Step-by-step explanation:

Answer:

attached below

Step-by-step explanation:

The current as  t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms

It is defined as the relationship between current and voltage according to the ohms law the voltage is directly proportional to the current.

It is given that:

A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms.

Using the equation:

LdI/dt + RI = v(t)

I(0) = 0

LdI/dt  = v(t)

V = 20 volts

After integration:

I = 2/7 + c

c = e⁻⁷⁰⁰ⁿ  (n = t) = 0

t → ∞

I = 2/7 A

Thus, the current as  t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms

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Answer:

Step-by-step explanation:

The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.

The detailed analysis and step by step approach is as shown in the attachment.

Answer:

1.66667

Step-by-step explanation:

You would do 5 divided by 3.

5 divided by 3 = 1.66667

You can verify this answer by multiplying 1.66667 * 3

1.66667 * 3 = 5

Hope this helps!!!

Answer:

Step-by-step explanation:

Given Data:

Set A :        x = 23,       Med = 22,             S = 1.2

Set B :        x = 24,       Med = 29,             S = 3.1

The Formula for Pearson's Index of Skewness (for Median in given data) is:

[tex]Sk_{2} = 3(\frac{x - Med}{S} )[/tex]

where,

[tex]Sk_{2} =[/tex] Pearson's Coefficient of Skewness

[tex]Med =[/tex] Median of Distribution

[tex]x=[/tex] Mean of Distribution

[tex]S=[/tex] Standard Deviation of Distribution

a) Finding Skewness:

For Set A:

[tex]Sk_{2_{A}} = 3(\frac{23 - 22}{1.2} )\\\\Sk_{2_{A}} = (\frac{3}{1.2} )\\\\Sk_{2_{A}} = 2.5[/tex]

For Set B:

[tex]Sk_{2_{B}} = 3(\frac{24 - 29}{3.1} )\\\\Sk_{2_{B}} = 3(\frac{-5}{3.1} )\\\\Sk_{2_{B}} = -4.84[/tex]

b) Type of Distribution:

For Set A:

As the value of skewness is a positive value (i.e. 2.5). Hence, Set A is right (positively) skewed.

For Set B:

As the value of skewness is a negative value (i.e. -4.84). Hence, Set B is skewed left (or negatively skewed).

c) Which Set can be considered as symmetric?

As the Pearson's Coefficient of skewness for Set A (2.5) is closer to 0 as compared to that of Set B (-4.84). Set A is more closer to that of a symmetric distribution and therefore can be considered as one.

Answer:

Is there a picture?

Step-by-step explanation:

It would be better if i could see a image!