Pierre sold 340 tickets for a concert. Balcony tickets cost $5 while tickets for the lower floor cost $10. If Pierre sold $2,700 worth of tickets, how many lower floor tickets did he sell?

Answer: 429

Step-by-step explanation:

If all but 1/13 of the students were involved in the evening's activities, that means that 12/13 of the students were involved.

To calculate the total number of students, first you do a simple rule of three to know the number of students that weren't involved in the activities:

If 12/13 of the students represent 396 students, then how many students are 1/13 of the students?

[tex]\frac{\frac{1}{13} *396}{\frac{12}{13} } =\frac{396}{12} =33[/tex]

Then, you add this number to the number of students that were involved in the activities:

[tex]396+33=429[/tex]

So, 429 is the number of students that attend the school.

You can also calculate it directly by doing the following rule of three:

If 12/13 of students represent 396 students, then how many students are 13/13 of the students?

[tex]\frac{\frac{13}{13} *396}{\frac{12}{13} } =\frac{396*13}{12} =\frac{5148}{12} =429[/tex]

Answer:429

Step-by-step explanation: let the total number of students in the elementary school be X

X - X/13 = 396

12X/13=396

X=396/*13/12

X=5148/12

X=429

The required probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.

To find the probability of rolling a dice and getting an even number (3, 4, or 6) and a number that is a multiple of 3 (3 or 6), we need to consider the outcomes that satisfy both conditions.

There are three even numbers on a standard six-sided dice: 2, 4, and 6. Out of these three even numbers, two of them are multiples of 3 (3 and 6).

The probability of rolling an even number is 3/6 (3 even numbers out of 6 possible outcomes) or simply 1/2.

The probability of rolling a number that is a multiple of 3 is 2/6 (2 multiples of 3 out of 6 possible outcomes) or 1/3.

Now, to find the probability of both events occurring together (rolling an even number and a number that is a multiple of 3), we multiply the individual probabilities:

Probability of getting an even number AND a number that is a multiple of 3 = (1/2) * (1/3) = 1/6.

So, the probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.

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The probability of rolling an even number is 1/2, while the probability of rolling a multiple of 3 is 1/3. However, the probability of rolling a number that is both an even number and a multiple of 3 (which is only the number 6) is 1/6 or approximately 16.67%.

The experiment in question involves rolling a single die and finding the probability of getting an even number as well as a number that is a multiple of 3. A standard die has six faces, numbered from 1 to 6. An even number on a die could be 2, 4, or 6, while a multiple of 3 could be 3 or 6.

To find the probability of getting an even number, we count the favorable outcomes for an even number, which are three (2, 4, 6), and divide by the total number of outcomes possible on a die, which is six. This gives us a probability of ½ or 50% for rolling an even number.

To find the number that is a multiple of 3, we have two outcomes (3 and 6). Again, dividing by the total number of outcomes we get a probability of ⅓ or approximately 33.33%.

But in this case, we are interested in the occurrence of both events together, which means we are looking for the probability of getting a number that is both an even number and a multiple of 3. The only number on a die that is both even and a multiple of 3 is the number 6.

Since there is only one favorable outcome (6) and six possible outcomes in total, the probability of rolling a number that is both even and a multiple of 3 is ⅖ or approximately 16.67%.

Answer with Step-by-step explanation:

Let A, B and C are arbitrary sets within a universal set U.

We  have to prove that [tex]( A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.

Let [tex](x,y)\in (A/B)\times C[/tex]

Then [tex] x\in(A/B) [/tex] and [tex] y\in C[/tex]

Therefore, [tex] x\in A[/tex] and [tex] x\notin B[/tex]

Then, (x,y) belongs to [tex] A\times C[/tex]

and (x,y) does not belongs to [tex] B\times C[/tex]

Hence,[tex] (x,y)\in(A\times C)/(B\times C)[/tex]

Conversely ,Let (x ,y)belongs to [tex] (A\times C)/(B\times C)[/tex]

Then [tex] (x,y)\in (A\times C)[/tex] and [tex] (x,y)\notin (B\times C)[/tex]

Therefore,[tex] x\in A,y\in C[/tex] and [tex] x\notin B,y\in C[/tex]

[tex] x\in(A/B)[/tex] and [tex]y\in C[/tex]

Hence, [tex] (x,y)\in(A/B)\times C[/tex]

Therefore,[tex] (A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.

Hence, proved.

Answer: Second Option

[tex]P(180<X <185)=13.5\%[/tex]

We know that the mean is:

[tex]\mu=175[/tex]

and the standard deviation is:

[tex]\sigma=5[/tex]

We are looking at the percentage of students between 180 centimeters and 185 centimeters in height.

This is:

[tex]P(180<X <185)[/tex]

We calculate the Z-score using the formula:

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

For [tex]X=180[/tex]

[tex]Z_{180}=\frac{180-175}{5}[/tex]

[tex]Z_{180}=1[/tex]

For [tex]X=185[/tex]

[tex]Z_{185}=\frac{185-175}{5}[/tex]

[tex]Z_{185}=2[/tex]

Then we look at the normal table

[tex]P(1<Z<2)[/tex]

[tex]P(1<Z<2)=P(Z<2)-P(Z<1)[/tex]

[tex]P(1<Z<2)=0.9772-0.8413[/tex]

[tex]P(1<Z<2)=0.135[/tex]

[tex]P(180<X <185)=13.5\%[/tex]

Note: You can get the same conclusion using the empirical rule

Look at the attached image for [tex]\mu+ 1\sigma <\mu <\mu + 2\sigma[/tex]

Answer:

the answer is Y=(6/17)X + (285/17)

Step-by-step explanation:

1. Identify the X and Y coordinate of each point.            

for example, the first point could be (12,21), therefore X1 = 12 and Y1 = 21 and the second point is then (46,33), therefore X2 = 46 and Y2 = 33

2. to find the equation of the line it is necessary to find the slope.

knowing that the equiation of the slope is M = Y2-Y1/X2-X1 we replace

  [tex]m=\frac{33-21}{46-12}=\frac{12}{34}[/tex] and then we simplify the result to [tex]\frac{12}{34}[/tex]

3. Using the Point-slope equation which is  Y-Y1=M(X-X1) we replace

 [tex]Y-33=\frac{6}{17}(X-46)[/tex]

[tex]Y=\frac{6}{17}X-\frac{276}{17}+33[/tex]

[tex]Y=\frac{6}{17}X+\frac{285}{17}[/tex]

Answer:

The probability that the chosen student is a junior is [tex]\frac{3}{10}[/tex]

Step-by-step explanation:

Given,

The total number of students = 40,

In which, freshmen = 15,

Sophomores = 8,

Juniors = 12,

And, seniors = 5,

Thus, the probability that junior student when a student is chosen

[tex]=\frac{\text{Total possible arrangement of 1 junior student }}{\text{Total possible arrangement of 1 student}}[/tex]

[tex]=\frac{^{12}C_1}{^{40}C_1}[/tex]

[tex]=\frac{12}{40}[/tex]

[tex]=\frac{3}{10}[/tex]

The probability of selecting a junior student is 0.3 or 30%.

To find the probability of selecting a junior student, we need to divide the number of junior students by the total number of students. There are 12 junior students out of a total of 40 students, so the probability can be calculated as:

Probability = Number of junior students / Total number of students

Probability = 12 / 40

Probability = 0.3

Therefore, the probability of selecting a junior student is 0.3 or 30%.

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Answer: 31,365

Step-by-step explanation:

Given : The number of horror films = 15

The number of comedies = 18

Then , the number of different combinations of 4 movies can he rent if he wants at least two comedies is given by :-

[tex]^{18}C_2\times ^{15}C_2+^{18}C_3\times ^{15}C_1+^{18}C_4\times ^{15}C_0\\\\=\dfrac{18!}{2!(18-2)!}\times\dfrac{15!}{2!(15-2)!}+\dfrac{18!}{3!(18-3)!}\times\dfrac{15!}{1!(15-1)!}+\dfrac{18!}{4!(18-4)!}\times\dfrac{15!}{0!(15-0)!}\\\\=16065+12240+3060=31365[/tex]

Hence, the  number of different combinations of 4 movies can he rent if he wants at least two comedies = 31,365

The student can calculate different combinations of 4 movies by considering the cases for two, three, or four comedies, and then adding up the combinations for each case.

The student is interested in renting movies and wants to know how many different combinations of 4 movies he can rent if he requires at least two comedies from a selection of 15 horror films and 18 comedies. We can use the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items to choose from, k is the number of items to choose, and ! denotes factorial.

To calculate the combinations with at least two comedies, we must consider the following cases:

For each case, we calculate the combinations separately for comedies and horror films and then multiply them:

Finally, we add all these possibilities together to get the total number of combinations.

Answer:

See explanation.

Step-by-step explanation:

The hyperbolic sine and cosine functions are defined as follows:

[tex] \sinh(x) = \frac{ {e}^{x} - {e}^{ - x} }{2} [/tex]

[tex]\cosh(x) = \frac{ {e}^{x} + {e}^{ - x} }{2} [/tex]

We want to show that:

[tex]\cosh^{2} (x) - \sinh^{2} (x) = 1[/tex]

We use the definition by substituting the expressions into the left hand side and simplify to obtain the RHS.

[tex] \cosh^{2} (x) - \sinh^{2} (x) = {( \frac{ {e}^{x} + {e}^{ - x} }{2} )}^{2} + {( \frac{ {e}^{x} - {e}^{ - x} }{2} )}^{2} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} + {e}^{ - 2x} - {e}^{2x} + {e}^{ - 2x }+ 2 +2 }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ 4 }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x)=1[/tex]

b)

If we start with the identity in a) and we divide both sides by cosh²x we get:

[tex] \frac{\cosh^{2} (x) }{\cosh^{2} (x) } -\frac{\sinh^{2} (x) }{\cosh^{2} (x) } =\frac{1}{\cosh^{2} (x) } [/tex]

This simplifies to:

[tex]1 - \tanh ^{2} (x) = \sec \: h ^{2} (x) [/tex]

Answer:

There are none.

Step-by-step explanation:

gcd(431,29)=1  

The reasons I came to this conclusion is because 29 is prime.

29 is not a factor of 431 so we are done.

So now we want to find (x,y) such that 115x+30y=1.

I'm going to use Euclidean's Algorithm.

115=30(3)+25

30=25(1)+5

25=5(5)

So we know we are done when we get the remainder is 0 and I like to look at the line before the remainder 0 line to see the greatest common divisor or 115 and 30 is 5.

So 115x+30y=5 has integer solutions (x,y) where d=5 is the smallest possible positive such that 115x+30y=d will have integer solutions (x,y).

So since 1 is smaller than 5 and we are trying to solve 115x+30y=1 for integer solutions (x,y), there there is none.

Furthermore,  115x+30y=1 can be written as 5(23x+6y)=1 and we know that 5 is not a factor of 1.

Answer:

The horizontal tangents occur at: (9,-2) and (11,2)

The vertical tangent occurs at (8.75,-1.375)

See attachment

Step-by-step explanation:

The given parametric equations are:

[tex]x=t^2-t+9[/tex] and [tex]y=t^3-3t[/tex]

The slope function is given by:

[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]

[tex]\frac{dy}{dx}=\frac{3t^2-3}{2t-1}[/tex]

The tangent is vertical when [tex]\frac{dx}{dt}=0[/tex]

[tex]\implies 2t-1=0[/tex]

[tex]t=\frac{1}{2}[/tex]

When [tex]t=\frac{1}{2}[/tex], [tex]x=(\frac{1}{2})^2-\frac{1}{2}+9=8.75[/tex], [tex]y=0.5^3-3(0.5)=-1.375[/tex]

The vertical tangent occurs at (8.75,-1.375)

The tangent is horizontal when [tex]\frac{dy}{dt}=0[/tex]

[tex]3t^2-3=0[/tex]

[tex]\implies t=\pm1[/tex]

When t=1, [tex]x=(1)^2-1+9=9[/tex], [tex]y=1^3-3(1)=-2[/tex]

When t=-1, [tex]x=(-1)^2+1+9=11[/tex], [tex]y=(-1)^3-3(-1)=2[/tex]

The horizontal tangents occur at: (9,-2) and (11,2)

Answer with explanation:

The given function in x and y is,

  y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

[tex]y'= -\csc^2 x+2 \csc x\times \cot x[/tex]

Slope of tangent at (π/2,3)

 [tex]y'_{(\frac{\pi}{2},3)}= -\csc^2\frac{\pi}{2} +2 \csc \frac{\pi}{2}\times \cot \frac{\pi}{2}\\\\=-1+2\times 1 \times 0\\\\= -1[/tex]

Equation of tangent passing through (π/2,3) can be obtained by

[tex]\rightarrow \frac{y-y_{1}}{x-x_{1}}=m(\text{Slope})\\\\ \rightarrow \frac{y-3}{x-\frac{\pi}{2}}=-1\\\\\rightarrow 3-y=x-\frac{\pi}{2}\\\\\rightarrow x+y-3-\frac{\pi}{2}=0[/tex]

⇒There will be no Horizontal tangent from the point (π/2,3).

Final answer:

The equation of the tangent to the curve y = 5 + cot x - 2 csc x at a given point can be found using the derivative, but for point P (π/2, 3), the slope is undefined because cot x and csc x are undefined at π/2. A horizontal tangent occurs where the derivative is zero, and point Q must be determined by solving y' = 0.

Explanation:

To find the equation of the tangent at point P, we first need to compute the derivative of the given function y = 5 + cot x - 2 csc x to find the slope of the tangent at any point on the curve. For the given curve y = 5 + cot x - 2 csc x, the derivative y' will give us the slope of the tangent at any point x. The slope of the tangent at point P, where P is (π/2, 3), can be found by evaluating the derivative at x = π/2.

However, since cot x and csc x are undefined at x = π/2, this curve will not have a well-defined slope and therefore not a conventional tangent line at x= π/2. On the other hand, a horizontal tangent will occur at points where the derivative y' is equal to zero. To find point Q, we must solve y' = 0 for x, and then use the x-value to find the corresponding y-coordinate on the curve.

Answer:

1. The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].

2. The value of a is 18.

Step-by-step explanation:

If a line passes through two points, then the slope of the line is

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

(1)

It is given that the line passes through the points (-a + 3, b - 3) and (a + 3, -b). So, the slope of the line is

[tex]m=\frac{-b-(b-3)}{a+3-(-a+3)}[/tex]

[tex]m=\frac{-b-b+3}{a+3+a-3)}[/tex]

[tex]m=\frac{-2b+3}{2a}[/tex]

The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].

(2)

If the line passing through the points (a, 1) and (6, 5), then the slope of the line is

[tex]m_1=\frac{5-1}{6-a}=\frac{4}{6-a}[/tex]

If the line passing through the points (2, 7) and (a + 2, 1), then the slope of the line is

[tex]m_2=\frac{1-7}{a+2-2}=\frac{-6}{a}[/tex]

The slopes of two parallel lines are same.

[tex]m_1=m_2[/tex]

[tex]\frac{4}{6-a}=\frac{-6}{a}[/tex]

On cross multiplication we get

[tex]4a=-6(6-a)[/tex]

[tex]4a=-36+6a[/tex]

[tex]4a-6a=-36[/tex]

[tex]-2a=-36[/tex]

Divide both sides by -2.

[tex]a=18[/tex]

Therefore the value of a is 18.

Answer:

Yes

Step-by-step explanation:

If I am understanding the question correctly, then I would have to say it is compared to "congruent".

Congruent: two triangles are compared and have the same sides and angles

Equilateral: all 3 sides of the triangle are the same.

Basing my opinion on the definitions above. I believe this situation could be compared to "congruent" because as mentioned in the question we are comparing ourselves to a triangle that has 3 DIFFERENT career choices.

If this was compared to "Equal" there would be 1 career choice repeated 3 times.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

During July, Pulam actually sold 42,000 units.

Variable Manufacturing Costs for 42,000 units  are-

1. Direct Material : [tex](36000/36000)\times42000 =42000[/tex] dollars

2. Direct Labor:  [tex](72000/36000)\times42000 =84000[/tex] dollars

3. Variable Manufacturing Overhead : [tex](72000/36000)\times42000 =84000[/tex] dollars

4. Fixed manufacturing costs : $ 180000

5. Total manufacturing costs : [tex]42000+84000+84000+180000=390000[/tex] dollars

Answer:

a) Marlon Brando

Step-by-step explanation:

Don Vito Corleone was played by Marlon Brando in the film Godfather released in the year 1972 directed by Francis Ford Coppola based on the 1969 book by Mario Puzo. Paramount studios was against the casting of Marlon Brando but the director insisted on giving him a screen test. The screen test convinced the studio to cast him in the role.

Marlon Brando starred as Don Vito Corleone in The Godfather, earning an Academy Award for his performance.

The established actor who starred in The Godfather as Don Vito Corleone is Marlon Brando. Brando's performance in this iconic film is widely recognized as one of his greatest roles. Released in 1972, The Godfather is a crime film directed by Francis Ford Coppola and based on the novel of the same name by Mario Puzo. Marlon Brando's portrayal of the mafia patriarch earned him an Academy Award for Best Actor, which further solidified his legacy as one of Hollywood's legendary actors. The other actors listed, John Wayne and Henry Fonda, were not part of the cast of The Godfather.

In order for [tex]p_{X,Y}(x,y)[/tex] to be a valid PMF, its integral over the distribution's support must be equal to 1.

[tex]p_{X,Y}(x,y)=\begin{cases}\dfrac{c(x+y)}2&\text{for }x\in\{1,2,4\}\text{ and }y\in\{1,3\}\\\\0&\text{otherwise}\end{cases}[/tex]

There are 3*2 = 6 possible outcomes for this distribution, so that

[tex]\displaystyle\sum_{x,y}p_{X,Y}(x,y)=\sum_{x\in\{1,2,4\}}\sum_{y\in\{1,3\}}\frac{c(x+y)}2=1[/tex]

[tex]1=\displaystyle\frac c2\sum_{x\in\{1,2,4\}}((x+1)+(x+3))=\frac c2\sum_{x\in\{1,2,4\}(2x+4)[/tex]

[tex]1=\displaystyle\frac c2((2+4)+(4+4)+(8+4))[/tex]

[tex]1=13c\implies\boxed{c=\dfrac1{13}}[/tex]

Answer:

[tex]4 \cdot(5 + 16) - 2 = 82[/tex]

Step-by-step explanation:

The given equation is

[tex]4 \cdot5 + 16 - 2 = 82[/tex]

We want to insert parenthesis, so that this equation will be true.

We insert parenthesis to obtain:

[tex]4 \cdot(5 + 16) - 2 [/tex]

We now use PEDMAS to simplify.

[tex]4 \cdot(21) - 2 =82[/tex]

[tex]84 - 2 = 82[/tex]

[tex]82 = 82[/tex]

Answer:

The function is equal to [tex]y=-(1/4)(x-8)^{2}-1[/tex]

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

[tex]y=a(x-h)^{2}+k[/tex]

where

a is a coefficient

(h,k) is the vertex

In this problem we have

(h,k)=(8,-1)

substitute

[tex]y=a(x-8)^{2}-1[/tex]

Find the value of a

Remember that we have the y-intercept

The y-intercept is the point (0,-17)

substitute

x=0,y=-17

[tex]-17=a(0-8)^{2}-1[/tex]

[tex]-17=64a-1[/tex]

[tex]64a=-17+1[/tex]

[tex]64a=-16[/tex]

[tex]a=-16/64[/tex]

[tex]a=-1/4[/tex]

therefore

The function is equal to

[tex]y=-(1/4)(x-8)^{2}-1[/tex]

see the attached figure to better understand the problem

Answer:

The correct option is C.

Step-by-step explanation:

If (x-c) is a factor of a polynomial f(x), then f(c)=0.

It is given that (x-1) is a factor of the polynomial. It means the value of the function at x=1 is 0.

In option A,

The given function is

[tex]p(x)=x^3+x^2-2x+1[/tex]

Substitute x=1 in the given function.

[tex]p(1)=(1)^3+(1)^2-2(1)+1=1+1-2+1=1[/tex]

Since p(1)≠0, therefore option A is incorrect.

In option B,

The given function is

[tex]q(x)=2x^3-x^2+x-1[/tex]

Substitute x=1 in the given function.

[tex]q(1)=2(1)^3-(1)^2+(1)-1=2-1+1-1=1[/tex]

Since q(1)≠0, therefore option B is incorrect.

In option C,

The given function is

[tex]r(x)=3x^3-x-2[/tex]

Substitute x=1 in the given function.

[tex]r(1)=3(1)^3-(1)-2=3-1-2=0[/tex]

Since r(1)=0, therefore option C is correct.

In option D,

The given function is

[tex]s(x)=-3x^3+3x+1[/tex]

Substitute x=1 in the given function.

[tex]s(1)=-3(1)^3+3(1)+1=-3+3+1=1[/tex]

Since s(1)≠0, therefore option D is incorrect.

Answer: Bill can make 24 different kinds of sandwiches

Step-by-step explanation:

Given : The number of kinds of rolls = 3

The number of varieties of meat = 4

The number of types of sliced cheese = 2

If he chooses one roll, one meat, and one type of cheese, then the number of different kinds of sandwiches he can make is given by :-

[tex]3\times2\times4=24[/tex]

Hence, Bill can make 24 different kinds of sandwiches.

Answer:

[tex](f+g)(x)=x-3+x^2[/tex] ; Domain = (-∞, ∞)

[tex](f-g)(x)=x-3-x^2[/tex] ; Domain = (-∞, ∞)

[tex](fg)(x)=x^3-3x^2[/tex] ; Domain = (-∞, ∞)

[tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex] ; Domain = (-∞,0)∪(0, ∞)

Step-by-step explanation:

The given functions are

[tex]f(x)=x-3[/tex]

[tex]g(x)=x^2[/tex]

1.

[tex](f+g)(x)=f(x)+g(x)[/tex]

Substitute the values of the given functions.

[tex](f+g)(x)=(x-3)+x^2[/tex]

[tex](f+g)(x)=x-3+x^2[/tex]

The function [tex](f+g)(x)=x-3+x^2[/tex] is a polynomial which is defined for all real values x.

Domain of (f+g)(x) = (-∞, ∞)

2.

[tex](f-g)(x)=f(x)-g(x)[/tex]

Substitute the values of the given functions.

[tex](f-g)(x)=(x-3)-x^2[/tex]

[tex](f-g)(x)=x-3-x^2[/tex]

The function [tex](f-g)(x)=x-3-x^2[/tex] is a polynomial which is defined for all real values x.

Domain of (f-g)(x) = (-∞, ∞)

3.

[tex](fg)(x)=f(x)g(x)[/tex]

Substitute the values of the given functions.

[tex](fg)(x)=(x-3)x^2[/tex]

[tex](fg)(x)=x^3-3x^2[/tex]

The function [tex](fg)(x)=x^3-3x^2[/tex] is a polynomial which is defined for all real values x.

Domain of (fg)(x) = (-∞, ∞)

4.

[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]

Substitute the values of the given functions.

[tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex]

The function [tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex] is a rational function which is defined for all real values x except 0.

Domain of (f/g)(x) = (-∞,0)∪(0, ∞)

[tex](f + g)(x) = x^2 + x - 3[/tex], domain: all real numbers.

[tex](f - g)(x) = -x^2 + x - 3[/tex], domain: all real numbers.

[tex](fg)(x) = x^3 - 3x^2[/tex], domain: all real numbers.

[tex]f(g(x)) = x^2 - 3[/tex], domain: all real numbers.

To find (f + g)(x), we need to add the functions f(x) and g(x).

The function f(x) = x - 3 and the function [tex]g(x) = x^2.[/tex]

So, [tex](f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).[/tex]

Expanding this equation, we get [tex](f + g)(x) = x^2 + x - 3.[/tex]

To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).

Since both f(x) = x - 3 and [tex]g(x) = x^2[/tex] are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.

To find (f - g)(x), we need to subtract the function g(x) from f(x).

So, [tex](f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).[/tex]

Expanding this equation, we get [tex](f - g)(x) = -x^2 + x - 3.[/tex]

The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find (fg)(x), we need to multiply the functions f(x) and g(x).

So, [tex](fg)(x) = f(x) * g(x) = (x - 3) * (x^2).[/tex]

Expanding this equation, we get [tex](fg)(x) = x^3 - 3x^2.[/tex]

The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find f(g(x)), we need to substitute g(x) into the function f(x).

So, [tex]f(g(x)) = f(x^2) = x^2 - 3.[/tex]

The domain of f(g(x)) is also all real numbers, as [tex]g(x) = x^2[/tex] is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.

In summary:

- [tex](f + g)(x) = x^2 + x - 3[/tex], domain: all real numbers.

- [tex](f - g)(x) = -x^2 + x - 3[/tex], domain: all real numbers.

- [tex](fg)(x) = x^3 - 3x^2[/tex], domain: all real numbers.

- [tex]f(g(x)) = x^2 - 3[/tex], domain: all real numbers.

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Answer:

Step-by-step explanation:

With the information you have provided, the volume can only be stated as a third degree polynomial.  We do not know the measurements of the square cut out of each corner, we can only call it "x".

If each side of the square measures 3 feet wide and we cut out 2 squares from a side, the length of that side is 3 - 2x.  Volume is length times width times height.  The length and the width will both be 3 - 2x, and the height is the measurement of x.  But since we don't know it, the height is just x.  Multiplying the length times the width times the height looks like this:

(3 - 2x)(3 - 2x)(x)

When you FOIL all this together you get a third degree polynomial

[tex]4x^3-12x^2+9x[/tex]

If you know the measurement of the squares cut out, plug that value in for x and get the volume in a number.

Answer:

Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%

Step-by-step explanation:

An investor can invest money and earn a stated interest rate = 15.40%

We have to find the nominal, periodic and effective interest rates for this investment.

Nominal rate = Stated interest rate is also called annual percentage rate or nominal interest rate. Therefore nominal interest rate = 15.40%.

Periodic interest rate = Periodic interest rate is the annual interest rate divided by number of compounding periods.

Periodic rate = ( [tex]\frac{15.40}{4}[/tex] ) = 3.85%

Effective interest rate = We have to calculate Effective interest rate (EAR)  by the formula =  [tex]EAR=[(\frac{1+i}{n})^{n}-1][/tex]

where i = rate of interest and n = number of compounding periods

[tex]EAR=[(1+\frac{0.154}{4})^4]-1[/tex]

[tex]EAR=[(1+.0385)^{4}]-1[/tex]

[tex]EAR=[(1.0385)^{4}]-1[/tex]

EAR = 1.1631 - 1

EAR = 0.1631 = 16.31%

Therefore, Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%

Answer: a) 91%  b) 9% c) 16%

Step-by-step explanation:

Let A be the event that homes for sale have​ garages and B be the event that homes for sale have​ swimming​ pools.

Now, given :[tex]P(A)=75\%=0.75[/tex]

[tex]P(B)=29\%=0.29[/tex]

[tex]P(A\cap B)=13\%=0.13[/tex]

a) [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow\ P(A\cup B)=0.75+0.29-0.13=0.91[/tex]

Hence, the probability that a home for sale has a pool or a​ garage is 91%.

b) The probability that a home for sale has ​neither a pool nor a​ garage is given by :-

[tex]1-P(A\cup B)=1-0.91=0.09[/tex]

Hence, the probability that a home for sale has ​neither a pool nor a​ garage is 9%.

c) The probability that a home for sale has a pool but no​ garage is given by :-

[tex]P(B)-P(A\cap B)=0.29-0.13=0.16[/tex]

Hence, the probability that a home for sale has a pool but no​ garage is 16%.

Final answer:

The probability a home for sale has a pool or garage is 91%, the probability of having neither feature is 9%, and the probability of having a pool but no garage is 16%.

Explanation:

The question involves solving problems related to probability and set theory. To find out the probability of events involving homes for sale with certain features, we can use the principle of inclusion and exclusion for probabilities.

Answering Part a)

The probability that a home has a pool or a garage is given by the formula P( A or B ) = P( A ) + P( B ) - P( A and B ). Thus, the probability is:

P( Pool or Garage ) = P( Pool ) + P( Garage ) - P( Pool and Garage )

= 0.29 + 0.75 - 0.13

= 0.91 or 91%

Answering Part b)

To find a home with neither a pool nor a garage, we subtract the probability of having either from 1. Therefore:

P( Neither ) = 1 - P( Pool or Garage )

= 1 - 0.91

= 0.09 or 9%

Answering Part c)

The probability of a home having a pool but no garage is calculated by taking the probability of having a pool and subtracting the probability of having both a pool and a garage:

P( Pool but no Garage ) = P( Pool ) - P( Pool and Garage )

= 0.29 - 0.13

= 0.16 or 16%

Answer:

Step-by-step explanation:

cost function

Cost for x tambs is the sum of start-up cost and per-unit cost multiplied by the number of units.

  c(x) = 18263 +1.14x

revenue function

Revenue from the sale of x tambs is the product of their price and the number sold.

  r(x) = 4.06x

profit function

Profit from the sale of x tambs is the difference between the revenue and cost:

  p(x) = r(x) -c(x)

  p(x) = 4.06x -(18263 +1.14x)

  p(x) = 2.92x -18263

The cost function, C(x), for the company is [tex]\( C(x) = 1.14x + 18263 \)[/tex] .The revenue function, R(x), for the company is [tex]\( R(x) = 4.06x[/tex] .The profit function, P(x), for the company is [tex]( P(x) = R(x) - C(x) = 4.06x - (1.14x + 18263) \)[/tex] .

To determine the profit function for the company, we need to calculate the total cost and total revenue for making and selling 'x' thing-a-ma-bobs, respectively, and then subtract the total cost from the total revenue.1. The cost function, C(x), is the sum of the variable cost (cost per unit times the number of units) and the fixed cost (start-up cost). Since it costs $1.14 to make each thing-a-ma-bob and the start-up cost is $18263, the cost function is:[tex]\[ C(x) = (\text{Cost per unit}) \times x + \text{Fixed cost} \]\[ C(x) = 1.14x + 182632[/tex]

. The revenue function, R(x), is the amount of money the company earns from selling 'x' thing-a-ma-bobs at $4.06 each. Therefore, the revenue function is[tex]:\[ R(x) = (\text{Selling price per unit}) \times x \][ R(x) = 4.06x3.[/tex]The profit function, P(x), is the revenue minus the cost. To find the profit for 'x' thing-a-ma-bobs, we subtract the cost function from the revenue function:[tex]\[ P(x) = R(x) - C(x) \]\[ P(x) = 4.06x - (1.14x + 18263) \]\[ P(x) = 4.06x - 1.14x - 18263 \]\[ P(x) = 2.92x - 18263 \]So, the profit function for the company is \( P(x) = 2.92x - 18263 \),[/tex]

where 'x' represents the number of thing-a-ma-bobs made and sold.

We're looking for a solution

[tex]y=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]

which has second derivative

[tex]y''=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/tex]

Substituting these into the ODE gives

[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^{n+2}=0[/tex]

[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]

[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]

[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty\bigg((n+2)(n+1)a_{n+2}-a_{n-2}\bigg)x^n=0[/tex]

Right away we see [tex]a_2=a_3=0[/tex], and the coefficients are given according to the recurrence

[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_2=0\\a_3=0\\n(n-1)a_n=a_{n-4}&\text{for }n\ge4\end{cases}[/tex]

There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.

[tex]k=0\implies n=0\implies a_0=a_0[/tex]

[tex]k=1\implies n=4\implies a_4=\dfrac{a_0}{4\cdot3}=\dfrac2{4!}a_0[/tex]

[tex]k=2\implies n=8\implies a_8=\dfrac{a_4}{8\cdot7}=\dfrac{6\cdot5\cdot2}{8!}a_0[/tex]

[tex]k=3\implies n=12\implies a_{12}=\dfrac{a_8}{12\cdot11}=\dfrac{10\cdot9\cdot6\cdot5\cdot2}{12!}a_0[/tex]

and so on, with the general pattern

[tex]a_{4k}=\dfrac{a_0}{(4k)!}\displaystyle\prod_{i=1}^k(4i-2)(4i-3)[/tex]

[tex]k=0\implies n=1\implies a_1=a_1[/tex]

[tex]k=1\implies n=5\implies a_5=\dfrac{a_1}{5\cdot4}=\dfrac{3\cdot2}{5!}a_1[/tex]

[tex]k=2\implies n=9\implies a_9=\dfrac{a_5}{9\cdot8}=\dfrac{7\cdot6\cdot3\cdot2}{9!}a_1[/tex]

[tex]k=3\implies n=13\implies a_{13}=\dfrac{a_9}{13\cdot12}=\dfrac{11\cdot10\cdot7\cdot6\cdot3\cdot2}{13!}a_1[/tex]

and so on, with

[tex]a_{4k+1}=\dfrac{a_1}{(4k+1)!}\displaystyle\prod_{i=1}^k(4i-1)(4i-2)[/tex]

[tex]a_2=0\implies a_6=a_{10}=\cdots=a_{4k+2}=0[/tex]

[tex]a_3=0\implies a_7=a_{11}=\cdots=a_{4k+3}=0[/tex]

Then the solution to this ODE is

[tex]\boxed{y(x)=\displaystyle\sum_{k=0}^\infty a_{4k}x^{4k}+\sum_{k=0}^\infty a_{4k+1}x^{4k+1}}[/tex]

Answer:

0.13774 inches ( approx ).

Step-by-step explanation:

Given,

The age of earth = 4.6 billion years = 4600000000 years,

The represented age of earth = 63,360 inches,

That is, scale factor in representation of age

[tex]=\frac{\text{Represented age}}{\text{Actual age}}[/tex]

[tex]=\frac{ 63,360}{4600000000}[/tex]

Now, the age of human civilization = 10000 years,

Thus, the represented age of human civilization = actual age × scale factor

[tex]=10000\times \frac{ 63,360}{4600000000}[/tex]

[tex]=\frac{633600000}{4600000000}[/tex]

[tex]=0.137739130435[/tex]

[tex]\approx 0.13774\text{ inches}[/tex]

Answer:

  about 15 hours

Step-by-step explanation:

You want to find t such that N(t)=200. Fill in the equation with that information and solve for t.

  200 = 400/(1 +399e^(-0.4t))

  1 +399e^(-0.4t) = 400/200 = 2 . . . . . multiply by (1+399e^(-0.4t))/200

  399e^(-0.4t) = 1 . . . . . . . . . . . . . . . . . . subtract 1

  e^(-0.4t) = 1/399 . . . . . . . . . . . . . . . . . .divide by 399

  -0.4t = ln(1/399) . . . . . . . take the natural log

  t = ln(399)/0.4 ≈ 14.972 . . . . . . . divide by -0.4, simplify

Rounded to tenths, it will take 15.0 hours for half the people to have heard the rumor.

The time it will take for half the group, or 200 people, to have heard the rumor based on the given exponential decay function is approximately 1.72 hours.

The problem presented is a mathematical question involving a model of exponential decay. In this context, we want to know when half of the people, or 200 people, will have heard the rumor. To find out, we substitute N(t) with 200 in the equation given:

200 = 400 / (1+399e^(-0.4t))

After simplifying and solving for t, t comes out to be approximately 1.72 hours. So, it takes approximately 1.72 hours for half the people to have heard the rumor.

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Final answer:

To express the confidence interval (0.023, 0.085) in the requested form, calculate the midpoint (p') as 0.054 and the margin of error (EBP) as 0.031. The confidence interval will thus be expressed as 0.054 - 0.031 < p < 0.054 + 0.031.

Explanation:

The question asks to express the confidence interval (0.023, 0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper E. To do this, we first need to calculate the midpoint (p') and the margin of error (EBP) of the given interval.

The midpoint (p') is the average of the lower and upper bounds of the interval, calculated as (0.023 + 0.085) / 2 = 0.054. The margin of error (EBP) is the difference between p' and either boundary of the interval, calculated as 0.085 - 0.054 = 0.031 or 0.054 - 0.023 = 0.031. Therefore, the interval in the requested format is 0.054 - 0.031 < p < 0.054 + 0.031.

The confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].

To express the confidence interval (0.023, 0.085) in the form [tex]\( \hat{p} - E < p < \hat{p} + E \),[/tex] where [tex]\( \hat{p} \)[/tex] is the point estimate and ( E ) is the margin of error, we need to find the point estimate and the margin of error.

Step 1: Find the point estimate [tex]\( \hat{p} \)[/tex].

The point estimate is the midpoint of the confidence interval. We can calculate it using the formula:

[tex]\[ \hat{p} = \frac{\text{Lower bound} + \text{Upper bound}}{2} \][/tex]

Substitute the values:

[tex]\[ \hat{p} = \frac{0.023 + 0.085}{2} \][/tex]

[tex]\[ \hat{p} = \frac{0.108}{2} \][/tex]

[tex]\[ \hat{p} = 0.054 \][/tex]

So, the point estimate [tex]\( \hat{p} \)[/tex] is 0.054.

Step 2: Find the margin of error \( E \).

The margin of error is half the width of the confidence interval. We can calculate it using the formula:

[tex]\[ E = \frac{\text{Upper bound} - \text{Lower bound}}{2} \][/tex]

Substitute the values:

[tex]\[ E = \frac{0.085 - 0.023}{2} \][/tex]

[tex]\[ E = \frac{0.062}{2} \][/tex]

[tex]\[ E = 0.031 \][/tex]

So, the margin of error ( E ) is 0.031.

Step 3: Express the confidence interval in the required form.

Now, we can write the confidence interval in the form [tex]\( \hat{p} - E < p < \hat{p} + E \):[/tex]

[tex]\[ 0.054 - 0.031 < p < 0.054 + 0.031 \][/tex]

[tex]\[ 0.023 < p < 0.085 \][/tex]

Therefore, the confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].

Answer:

unpaid balance is $156439.86

Step-by-step explanation:

present value= $190000

rate = 3% monthly = 3/12 = 0.25 % = 0.0025

time = 25 year  = 25 × 12 300

to find out

the unpaid balance

solution

first we calculate monthly payment

payment  = principal ( 1- (1+r)^(-t) /r )

and we put all value here

190000 = principal ( 1- (1+0.0025)^(-300) / 0.0025 )

principal = 190000 /  ( 1- (1+0.0025)^(-300) / 0.0025 )

principal = 901

so after 6 year unpaid is

present value  = principal ( 1- (1+r)^(-t) /r )

put here principal 901 and rate 0.0025 and time 300 - (year )

time = 300 - (6*12) = 300 - 96 = 228

payment   = 901 ( 1- (1+0.0025)^(-228) / 0.0025 )

payment  =  901 ( 713.63)

so unpaid balance is $156439.86