which of the following is always true for all probability density functions of continuous random variables? group of answer choices they are symmetrical they are bell-shaped the area under the curve is 1.0 they have the same height

Answer:

$1,300

Step-by-step explanation:

Let original price be $x

Computer has been marked down by 45%

45% = 45/100 = 0.45

So discount = 45% of x = 0.45 x

Original price - Discount = Sale price
x - 0.45x = $715

0.55x = 715

x = 715/0.55

x = $1,300

which is the original price

Answer:

Step-by-step explanation:

If James gave each cousin 3 cupcakes, there would be 25 left. James is 71 cupcakes short of being able to give each cousin 15 cupcakes. You want to know the number of cousins and the number of cupcakes.

The given relations between the number of cupcakes (n) and the number of cousins (c) are ...

  n -3c = 25 . . . . . 3 each ⇒ 25 left

  n -15c = -71 . . . . . 15 each ⇒ 71 short

Subtracting the second equation from the first, we have ...

  (n -3c) -(n -15c) = (25) -(-71)

  12c = 96

  c = 8 . . . . . . divide by 12

  n = 3c +25 = 24 +25 = 49

(a) James has 8 cousins.

(b) James baked 49 cupcakes.

A + B  =0  => {u + v, au} is linearly independent set of two vectors

=>  {u + v, au}is a basis for V.

Let u and v be distinct vectors of a vector space V and {u, v} is a basis for V and a and b are nonzero scalars.

We have to show: both {u + v, au} and {au, bv} are also bases for V.

[tex]u\neq v,a,b\neq 0[/tex]

{u, v}  is a basis for V => {u + v , au} and {au, bu} are bases for V

From basis {u , v} we get dimension of V

{u , v} is a basis for V

=>  V is a 2-dimensional vector space and (.) [tex]\alpha u+\beta v=0= > \alpha =\beta =0[/tex]

Show  {u + v, au}  is a basis for V:

{u + v, au} :

A(u + v) + B(au) = 0 <=> Au + Av + Bau = 0

<=> (A + Ba)u + (B)v = 0

By the step (in second step)

A + Ba = 0 , [tex]a\neq 0[/tex]

B = 0

=> A + B  =0  => {u + v, au} is linearly independent set of two vectors

=>  {u + v, au}is a basis for V.

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The proof of LMNO is a rhombus is shown below.

A parallelogram is a particular instance of a rhombus. The opposing sides and angles in a rhombus are parallel and equal. A rhombus also has equal-length sides on each side, and its diagonals meet at right angles to form its shape. The rhombus is also referred to as a diamond or rhombus.

Given:

|LO|=|MN| and |LM|=|ON|

Since Opposite sides of a parallelogram are equal.

Now, LN⊥OM

So, ∠LPO = ∠NPO = 90° ( by definition of perpendicular lines)

LPO ≅ ∠NPO (by definition of congruent angles)

|LP|=|PN| (diagonals of a parallelogram bisect each other)

Thus, LMNO is a rhombus

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Answer:

♣: ✔ All right angles are congruent.

♦: ✔ reflexive property

♠: ✔ Opposite sides of a parallelogram are congruent.

The probability that a newly arrived email which does mention "free gift card", is not spam is 0.0165.

Let S be the event that an email is spam, and let F be the event that an email mentions "free gift card". We want to find P(S' | F), the probability that the email is not spam given that it mentions "free gift card".

We can use Bayes' theorem to find P(S' | F):

P(S' | F) = P(F | S') P(S') / P(F)

We can calculate each of these probabilities as follows:

P(F | S') = the probability that a non-spam email mentions "free gift card". From the problem, we know that this probability is 0.005 (i.e., 99.5% of non-spam emails do not mention "free gift card").

P(S') = the probability that an email is not spam. From the problem, we know that this probability is 0.4 (i.e., 60% of emails are spam, so 40% are not spam).

P(F) = the probability that an email mentions "free gift card". This can be calculated using the law of total probability:

P(F) = P(F | S) P(S) + P(F | S') P(S')

= 0.20 × 0.60 + 0.005 × 0.40

= 0.121

In the first term, we use the given probability that 20% of spam emails mention "free gift card". In the second term, we use the probability that 0.5% of non-spam emails mention "free gift card".

Plugging these values into Bayes' theorem, we get:

P(S' | F) = 0.005 × 0.4 / 0.121 ≈ 0.0165

Therefore, the probability that a newly arrived email which does mention "free gift card" is not spam is approximately 0.0165 or 1.65%.

Correct Question :

A spam filer is designed by screening commonly occurring phrases in spam. Suppose that 60% of email is spam. In 20% of the spam emails, the phrase “free gift card” is used. In non-spam emails, 99.5% of them do not mention “free gift card”. What is the probability that a newly arrived email which does mention “free gift card”, is not spam?

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Answer:

29

Step-by-step explanation:

trust

Answer: okay good!!!!!!

By rewriting the equation in slope-intercept form and setting Celsius to 0, we can find the y-intercept of the corresponding line, which represents the temperature in Kelvin when Celsius is 0°C.

When studying linear relationships between two variables, it is often useful to identify the y-intercept of the line that represents such relationship.

The equation K = °C + 273.15 expresses the relationship between the temperature in Celsius (°C) and the temperature in Kelvin (K). It tells us that to convert a temperature from Celsius to Kelvin, we need to add 273.15 to the original value.

We can plot the Celsius and Kelvin values as two variables on a graph, with Celsius on the x-axis and Kelvin on the y-axis. Since the equation K = °C + 273.15 is a linear equation (meaning that the graph is a straight line), we can use the slope-intercept form of the equation to find the y-intercept, which is the point where the line intersects the y-axis.

The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept. In the case of K = °C + 273.15, we can rewrite it as K = 1°C + 273.15, which tells us that the slope of the line is 1. This means that for every 1°C increase in temperature, there is a corresponding 1 K increase in temperature.

To find the y-intercept, we can simply set the value of Celsius to 0, since this is the point where the line intersects the y-axis. Plugging in 0 for °C in the equation K = °C + 273.15, we get K = 273.15. This means that the y-intercept of the line is (0, 273.15), which corresponds to the temperature in Kelvin when the temperature in Celsius is 0°C.

In summary, the equation K = °C + 273.15 provides a linear relationship between the temperature in Celsius and the temperature in Kelvin.

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Answer: If the rate at which the flu spreads is proportional to the number of interactions between students who have the flu and students who have not been exposed to it, we can write:

dx/dt = kx(1000 - x)

where x is the number of students who have contracted the flu, k is the constant of proportionality, and t is time. The term x(1000 - x) represents the number of interactions between the students who have the flu and the students who have not been exposed to it. The factor of 1000 - x represents the number of students who have not yet been exposed to the flu.

Thus, the differential equation models the spread of the flu on the isolated college campus as the number of people who have contracted the flu changes over time due to the number of interactions between the students who have the flu and the students who have not been exposed to it.

Step-by-step explanation:

a) To compare the overall percentage of late deliveries for each delivery service, we need to calculate the total number of late packages and total number of deliveries for each service, and then divide the number of late packages by the total number of deliveries, and multiply by 100 to get the percentage.

For Pack Rats, the total number of deliveries is 400 + 100 = 500, and the total number of late packages is 12 + 16 = 28. Therefore, the percentage of late deliveries for Pack Rats is:

percentage of late deliveries for Pack Rats = (28/500) x 100 = 5.6%

percentage of late deliveries for Boxes R Us = (30/500) x 100 = 6%

Therefore, based on these calculations, Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us.

b) While Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us, it's important to note that this decision should not be based solely on this one comparison. It's possible that there are other factors that the company needs to consider, such as the cost of each delivery service or the quality of customer service provided by each company.

Furthermore, the sample size in this trial period may not be large enough to draw a definitive conclusion about the reliability of each service. Therefore, while Pack Rats may be a good choice based on the available data, the company should consider other factors and conduct further research before making a final decision.

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The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63​ = 12x+4y+51.

Simplify simply means to make it simple. In mathematics, simply or simplification is reducing the expression/fraction/problem in a simpler form. It makes the problem easy with calculations and solving.

here, we have,

now, we have to solve the given expression, using addition,

given that,

7x - 31 + 4y + 27 + 5x - 8 + 63​

=12x+4y+51

here, we add the terms with x with each other, then add the terms with y, and, add the numbers & subtract them from given terms with minus sign.

Hence, The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63​ = 12x+4y+51.

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Part A: the system of inequalities has a definite solution to it.

Part B: Yes the point is a part of Solution Area

What is Linear Inequality?

In mathematics a linear inequality is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality. It shows the data which is not equal in graph form.

Solution:

The inequalities 4x + 2y ≤ 16

And x + y ≥ 4 is attached below please refer to it.

Part A: The graph of both inequalities intersects each other therefore the system of inequalities has a definite solution to it.

Part B: Yes point (2, 3) is included in the solution area, the following is the mathematical proof.

Using substitution 4*2 + 2*3 ≤ 16

8+6 ≤ 16

14 ≤ 16

And 2+3 ≥ 4

5 ≥ 4

Hence, the point is included in the solution area.

Part C:

Let’s assume a point 1,1 in the solution set so it implies that only the inequality 4x + 2y ≤ 16 is satisfied. Thus, not giving a feasible solution.

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a) The probability of selecting a subject with a positive test result given that the subject does not use drugs is equals to the 0.09473.

b) The probability of selecting a subject with a negative test result given that the subject uses drugs is equals to the 0.12.

The confusion matrix of the pre employment drug screening data is present above. It contains positive test result, Negative test result and Total.

Subject uses drugs 44 6 = 50

Subject does not use drugs 90 860 = 950

Similarly , 90 + 44 = 134 , 860 + 6 = 886 and 950 + 50 = 1000.

a) We have to determine the probability of selecting a subject with a positive test result that the subject does not use drug. Let A be the event that the subject with a positive test result and B be the event that subject does not use drug. We have to determine P(A/B). From conditional probability, P(A/B) = P(A and B)/P(B)

= 90/950 = 0.09473

Hence, the probability of selecting a subject with a positive test result given that the subject does not use drug is 0.09473.

b) We have to calculate the probability of selecting a subject with a negative test result that the subject use drugs. Let R be the event that the subject with a negative test result and S be the event that subject uses drug. We have to calculate the value of P(R/S). By using Bayes theorem, P(R/S) = P(R and S) /P(S)

= 6/50 = 0.12

Hence, the probability of selecting a subject with a negative test result given that the subject uses drug is 0.12.

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Complete question:

The above table complete the question.

2. this data came from a pre-employment drug screening. positive test result negative test result (drug use is indicated) (drug use is not indicated) subject uses drugs 44 6 (true positive) (false negative) subject is not a drug user 90 860 (false positive) (true negative) (a) false positive find the probability of selecting a subject with a positive test result given that the subject does not use drugs. (b) false negative find the probability of selecting a subject with a negative test result given that the subject uses drugs.

Using relational algebra:

a. flight ⋈ airport ⋈ ρ destination  airport

b. flight ⋈ airport

c. passenger ⋈ airport

d. airline ⋈ airport ⋈ flight

e. airline ⋈ airport ⋈ flight

f. airline ⋈ airport ⋈ flight ⋈ ρ destination airport

g. flight ⋈ passenger ⋈ booking

QUESTION (a):

- using EQUI join (⋈) join relations "Flight, Airport and Airport as destination"

- and using select Operation (σ), relational operators (=, >) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "Flight.fID, Airport.code, destination.code, Flight.fare" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π flight . fid, airport . code, destination . code , flight.fare   σ flight . airport = airport . code

    AND flight . destination = destination . code  (flight ⋈ airport ⋈ ρ destination  airport)

QUESTION (b):

- using EQUI join (⋈) join relations "Flight, Airport"

- and using select Operation (σ), relational operators (=, >=, <=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airport . city, airport . code, flight . departure" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airport . city, airport . code, flight . departure    σ flight . airport = airport . code

     AND (4 <= flight . departure AND flight . departure <= 10)  (flight ⋈ airport)

QUESTION (c):

- using EQUI join (⋈) join relations "passenger, Airport"

- and using select Operation (σ), relational operators (<>) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, passenger . hometown" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, passenger . hometown    σ passenger . hometown <> airport . city  (passenger ⋈ airport)

QUESTION (d):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND airport . city = "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (e):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

     AND flight . airport = airport . code

    AND airport . city <> "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (f):

- using EQUI join (⋈) join relations "airline, airport, flight and airport as destination"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND flight . destination = destination . code

    AND airport . city = "Toronto"

    AND destination . city = "vancouver"  (airline ⋈ airport ⋈ flight ⋈ ρ destination airport)

QUESTION (g):

- using EQUI join (⋈) join relations "flight, passenger, Booking"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, airport, destination" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, airport, destination    σ passenger . pid = booking . pid

    AND booking . fid = flight . fid

    AND destination . city = "vancouver"  (flight ⋈ passenger ⋈ booking)

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Question correction:

See on the attached image.

In the linear equation, The value of the x and y are 11/6, -5 and  18/4.

The definition of a linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

here, we have,

We have to find that the value of the x and the y.

From the given information:

The equations are :

(4y - 18)

(6x - 11)

(x + 5)​

Now take all equation equal to zero.

Then

4y - 18 = 0

y = 18/4

And then

6x - 11 = 0

x = 11/6

And then

x + 5 = 0

x = -5.

So, In the linear equation, The value of the x and y are 11/6, -5 and  18/4.

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The option which is not a requirement for the permutation rule is d) Order is not taken into account (rearrangements of the same items are considered to be the same).

Permutations is the number of arrangements of r objects taken out of n objects.

The requirements for permutation include the folowing

Since these are the requirements for a permutation, the options a, b and c are requirements for a permutation while option d is not.

So, the option which is not a requirement is d) Order is not taken into account (rearrangements of the same items are considered to be the same).

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The problem involves finding a constant c that makes each given function a valid probability mass function given are f(x) = x/c, x=1,2,3,4; f(x) = cx, x=1,2,3,...,10; f(x) = c(x+1)², x=0,1,2,3; f(x) = x/c, x=1,2,3,...,n; f(x) = (x+1)(x+2)/[x(x+3)], x=0,1,2,3,...

(a) We know that the sum of all the probabilities in a pmf must be equal to 1.

Therefore, we have:

1/c + 2/c + 3/c + 4/c = 1

Solving for c, we get:

c = 10

The pmf can be depicted as:

x f(x)

1 0.1

2 0.2

3 0.3

4 0.4

(b) Again, we know that the sum of all the probabilities in a pmf must be equal to 1. Therefore, we have:

c(1 + 2 + 3 + ... + 10) = 1

Solving for c, we get:

c = 1/55

The pmf can be depicted as:

x f(x)

1 1/55

2 2/55

3 3/55

... ...

10 10/55

(d) We can use the same approach as in part (a) to find c:

c(1 + 4 + 9) = 1

Solving for c, we get:

c = 1/14

The pmf can be depicted as:

x f(x)

0 1/14

1 1/6

2 1/3

3 4/14

(e) Similar to part (a), we have:

1/c + 2/c + 3/c + ... + n/c = 1

Solving for c, we get:

c = (n+1)n/2

The pmf can be depicted as:

x f(x)

1 1/(n+1)

2 2/(n+1)

3 3/(n+1)

n n/(n+1)

(f) We can use the hint given in the problem:

f(x) = 1/(x + 1) - 1/(x + 2)

We can see that this is a telescoping series, where all terms cancel out except for the first and the last term. Therefore, we have:

f(0) = 1/1 - 1/2 = 1/2

f(1) = 1/2 - 1/3 = 1/6

f(2) = 1/3 - 1/4 = 1/12

f(3) = 1/4 - 1/5 = 1/20

The pmf can be depicted as:

x f(x)

0 1/2

1 1/6

2 1/12

3 1/20

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The number of calories burned per minute be 23.33333.

Equations simultaneously, or a system of equations Several equations in algebra must be solved concurrently (i.e., the solution must satisfy all the equations in the system). There must be an equal number of equations and unknowns for a system to have a singular solution.

In order to locate the point where the lines intersect when the equations are graphed, systems of equations must be solved. The (x,y) ordered pair of this intersection point is regarded as the system's solution.

Let j be the number of calories burned by Shelly while running and c be the number of calories she burns while cycling.

The system of equations be

45j + 30c = 1350

30j + 45c = 1350

45j + 30(10)=1350

j = 23.33333

Therefore, the value j be 23.33333.

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When the mean is higher than the median, the distribution has a right-skewed pattern. Thus the statement 'a. the distribution is positively skewed' is the accurate one.

This indicates that the majority of the data are centered on the left and that the right is covered by a lengthy tail. The mean exceeds the median. In a positively skewed distribution, the mean overestimates the values that are most prevalent. The median is larger than the mean. In a distribution that is positively skewed, the mean—the average of all the values—is higher than the median since the data tends to be lower. The center value of the data is called the median, in comparison. As a result, if the data is more skewed towards the negative, the average will be higher than the middle number.

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We cannot deduce anything about the p-value for age in the regression summary from just the 90% and 95%. cis for the parameter associated with age.

The 90% and 95% cis for the parameter associated with age tell us the range of values that the parameter could take, given the data. This range does not tell us anything about how likely the parameter is to take any of these values.

To determine how likely it is to take any of these values, we need to look at the p-value for age in the regression summary. The p-value tells us the probability that the parameter associated with age is different from zero.

It is the level of significance and is used to determine if the parameter is statistically significant. It is the only way to determine the likelihood of the parameter taking any of the values in its 90% and 95% cis.

Therefore, we cannot deduce anything about the p-value for age in the regression summary from just the 90% and 95% cis for the parameter associated with age.

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The length which lies from K Street between C Street and D Street is 337.5 ft.

Given that,

E Street, W Street, C Street, and D Street are parallel streets that cross K Street and M Street in a particular city. It is unknown how far K Street is from C Street and D Street.

Let the distance of street K between C and D be x,

Now,

Taking the equality of the proportionality expression of triangles,

600 / 400 = 600 + x / 400 + 250

6 / 4 = 600 + x / 625

3750/4 = 600 + x

937.5 = 600 + x

x = 337.5 ft

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A compound event can be described for flipping a coin two times and getting at least 1 head.

Compound event is defined as the events in which there are more than one event happens together.

All the outcomes of the events sum up the probability to 1.

We have to describe a compound event where the probability is in between 50% and 80%.

That is, probability is in between 0.5 and 0.8.

That is, there is more than half of the chance to occur the event.

Suppose that we flip a coin two times.

Sample space = {HH, HT, TH, TT}

Find the probability of getting at least 1 head.

There are 3 outcomes out of 4 of getting at least 1 head.

Probability = 3/4 = 75%

Hence there is a 75% probability for getting at least 1 head when flipped a coin two times.

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Answer:

a) Prime factorization of 396: 2² × 3¹ × 11¹

b) Factors of 396: 88 and 9.

Step-by-step explanation:

a) To find the prime factorization of 396, we can start by dividing it by the smallest prime number, which is 2:

396 ÷ 2 = 198

Now we can write:

396 = 2 × 198

Next, we can divide 198 by 2 again:

198 ÷ 2 = 99

So we have:

396 = 2 × 2 × 99

Now we need to find the prime factorization of 99. We can start by dividing it by 3:

99 ÷ 3 = 33

So we can write:

396 = 2 × 2 × 3 × 33

Now we need to find the prime factorization of 33. We can start by dividing it by 3 again:

33 ÷ 3 = 11

So we have:

396 = 2 × 2 × 3 × 11 × 1

We don't need to include the factor of 1, so we can write the prime factorization of 396 in index form as:

396 = 2² × 3¹ × 11¹

b) To find which two of the given numbers are factors of 396, we need to check if all the prime factors of each number are also prime factors of 396.

The prime factorization of 88 is 2³ × 11. Both 2 and 11 are factors of 396, so 88 is a factor of 396.

The prime factorization of 121 is 11². 11 is a factor of 396, but there is no factor of 121 that matches the remaining prime factors of 396 (2² × 3¹), so 121 is not a factor of 396.

The prime factorization of 9 is 3². Both 3 and 2 are factors of 396, so 9 is a factor of 396.

The prime factorization of 14 is 2 × 7. 7 is not a factor of 396, so 14 is not a factor of 396.

The prime factorization of 22 is 2 × 11. Both 2 and 11 are factors of 396, so 22 is a factor of 396.

Therefore, the two numbers that are factors of 396 are 88 and 9.

For the function for the temperature H = f(t) , for cup of coffee placed on the Kitchen counter , then the derivative f'(t) will be Negative .

The Temperature Function in degree Celsius for a cup of coffee placed on the kitchen counter will be negative because if cup of coffee is placed on the counter there is a loss of heat from cup of coffee to surroundings as temperature increases .

If the cup of coffee is losing heat to the surroundings, the temperature of the coffee will decrease over time.

The derivative of the temperature function, f'(t), will be negative. it means that as time increases, the rate of change of the temperature will be negative, indicating a decrease in temperature.

Therefore , the derivative function f'(t) is negative .

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The given question is incomplete , the complete question is

The temperature, H , in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by H = f(t) , where t is in minutes since the coffee was put on the counter .

Is f'(t) positive or negative ?

2.5 hours spent in each car wash would cost $40

From the question, we have the following parameters that can be used in our computation:

y = 10x + 20

y = 15x + 5

A situation for the system of equations is the total charges of different car wash per hour

So, the solution would be

The number of hours spent in each car wash for the total amount to be the same

From the graph, the solution is (2.5, 40)

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The integral for the volume of the solid is obtained by rotating the region bounded by the given curves about the specified axis is  V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

When we rotate a thin vertical strip, about the y-axis.

We get a cylindrical shell  with an inner of radius an x and an outer of radius x + dx

The height of the cylinder shell is tan (x) - x

The volume of the cylindrical shell is

     dV = π [tex](Outer Radius)^{2} (Height)[/tex] - π[tex](Inner Radius)^{2} (Height)[/tex]

     dV = π [tex](x + dx )^{2} (tan(x) - x)[/tex] - π [tex](x )^{2} (tan(x) - x)[/tex]

dV = π[tex](x^{2} + 2 xdx + (dx)^{2} ) (tan(x) - x)[/tex] - π [tex](x)^{2} (tan(x) - x)[/tex]

assume [tex]dx^{2}[/tex]≈ 0

            dV = π[tex](x^{2} + 2xdx + 0 - x^{2} ) (tan(x) - x)\\[/tex]

                    dV = 2πx (tan(x) - x) dx

                   V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

Therefore, the integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis is  

V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex] .

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The correct question is:

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=tanx,y=x,x=π/3; about the y-axis

Hence, The value of y = 12 when x = -6.

Given that , Y varies directly with x .

If y=−4 when x=2,

Find y when x=−6.

According to the question,

This is a direct variation , y = kx

If,

y = −4  when x=2,

y = kx

-4 = 2k

k = -2

To find ;

y when x = −6

y = kx

y = (-2) (-6)

y = 12

Out of the following the statement  has a reference angle of 30° and is in quadrant one or two-  is true of the location of the terminal side of an angle 0 whose sine value is 1/2.

Since the reason behind this is that the sine value of 1/2  can be written as sin(30°), this means that the terminal side of the angle is located at 30°, which is either quadrant one or two. Quadrant one is the upper right quadrant of the coordinate plane, and quadrant two is the upper left quadrant.

The reference angle of an angle is the angle between the terminal side of the angle and the positive x-axis. In this case, the reference angle would be  30°. The reference angle of an angle is important because it is used to determine the quadrant in which the terminal side of the angle is used to determine the quadrant in which the terminal side of the angle is located. The reference angle must be less than 90° in order for the terminal side to be located in the first and second quadrants.

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The moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex]

In probability theory, the negative binomial distribution is a discrete probability distribution that describes the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs.

The moment generating function (MGF) of a random variable X is defined as

[tex]M_X(t) = E(e^{tX})[/tex]

To derive the MGF of a negative binomial distribution, we can use the fact that a negative binomial random variable can be expressed as a sum of independent geometric random variables.

Let X ~ NB(r, p) be a negative binomial random variable, where r is the number of failures and p is the probability of success in each trial. Let Y ~ Geo(p) be a geometric random variable representing the number of failures before the first success.

We know that the MGF of a geometric distribution is

[tex]M_Y(t) = E(e^{tY}) = (1-p)/(1-pe^t).[/tex]

Using the fact that a negative binomial random variable is a sum of r independent geometric random variables, we can derive the MGF of X as follows:

[tex]M_X(t) = E(e^{tX}) = E(e^{tY_1 + tY_2 + ... + tY_r})\\\\= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]

(since the Y's are independent)

[tex]= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]

(since the Y's have the same distribution)

[tex]= [M_Y(t)]^r\\\\= [(1-p)/(1-pe^t)]^r[/tex]

Therefore, the moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex] This function can be used to derive moments and other statistical properties of the negative binomial distribution.

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