construct a boxplot for the following data and comment on the shape of the distribution representing the number of

Answer:

Step-by-step explanation:

For first game PA = 1/3

For second game PA = 1/3 ( If A is not lost in first game )

 = 2/5 (If A is lost in first game )

For conclusion of game in three matches :

If A wins , the probability is

PA , PB , PA  = 1/3 X 2/3 X 3/5 = 6/45

PB , PA , PA = 2/3 X 3/5 X 1/3 = 6/45

If B wins

PA, PB, PB = 1/3 X 2/3 X 2/5 =  4/ 45

PB, PA , PB = 2/3 X 3/5 X 2/3 = 4 / 15

Total probability of conclusion of game in 3 matches

= 6/45 +6/45 + 4/45 +4/15 = 28/45

b )

For the game concluding in 2 matches , the probability are as follows

PA,PA = 1/3 X 1/3 = 1/9

PBPB = 2/3 X 2/5 = 4 / 15

Total probability

= 1/9 + 4/15 = 17/45

So PA = 1/9 / 17/45

= 5/17

Answer:  10

Step-by-step explanation:

Given : Jason has five coins in his pocket: a penny, a nickel, a dime, a quarter, and a half-dollar.

Since each coin has different value from others.

So the combination of any 3 coin will give a different amount.

We know that the combination of r things out of n things = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

Therefore , the combination of 3 coins out of 5 = [tex]^5C_3=\dfrac{5!}{3!(5-3)!}=\dfrac{5\times4\times3!}{3!\times2!}=10[/tex]

Hence, the number of different sums of money can be formed using exactly three of the coins = 10

Answer:

Please refer to the attachment below.

Step-by-step explanation:

Please refer to the attachment below for explanation.

Answer:

Var(u​|X​=1) = 1

Var(Y|X​=1) = 1

​Var(u|X​) = 3

Var(Y|X​) = 9

Step-by-step explanation:

First, we need to identify the properties of the variance, so:

1. Var(a) = 0 , if a is constant

2. [tex]Var(aX) = a^{2} V(X)[/tex],  Where a is constant and X is a random variable

3. [tex]Var(aX+b)=a^{2} Var(X)+0[/tex], Where a and b are constants and X is a random variable.

4. [tex]Var(X + Y)=Var(X) + Var(Y)[/tex], Where X and Y are random variables and are independents.

Then, if ​Y=1​+X+u​, where X​, Y​, and ​u=v+X are random​ variables, v is independent of X​; ​E(v​)=0, ​Var(v​)=1​, ​E(X​)=1, and ​Var(X​)=2, the variance of the following cases are calculated as:

Var(u|X​=1) = Var( v + X | X = 1) = Var( v + 1 )    

                 = Var(v) + Var(1)                            

                  = 1 + 0 = 1

Var(Y|X​=1) = Var( 1 + X + u | X = 1 )

                 = Var (1 + X + v + X|X=1)

                 = Var(1 + 1 + v + 1)                      

                 = Var(3) + Var(v)                      

                 = 0 + 1 = 1

​Var(u|X​) = Var ( v + X | X)

             = Var ( v + X)

             = Var(v) + Var(X)

             =  1 + 2  = 3

Var(Y|X​) = Var ( 1 + X + u | X)

             = Var ( 1 + X + v + X | X)

             = Var ( 1 + 2X + v)

             = Var(1) + Var(2X) + Var(v)

              [tex]= Var(1)+2^{2}*Var(X)+Var(v)[/tex]

             = 0 + 4*2 + 1 = 8 + 1 = 9

Answer: [tex]a+b+c=\frac{n+1}{n}.[/tex]

Step-by-step explanation: The function [tex]f(x_1,x_2,\ldots)=x_1^2+x_2^2+\ldots[/tex] is always positive except at the origin where it is equal to zero. This means that the absolute minumum of this function must be [tex]a=0[/tex]. Absolute maximum is when all of the variables are equal to zero except [tex]x_1[/tex] which is equal to 1 (f evaluated at this point is equal to 1 do b=1). The function itself is then equal to 1. This is because when [tex]f(\cdots)=x_1^2+x_2^2+\ldots\leq x_1^2+2x_2^2+3x_3^2+\ldots\leq1[/tex] so it is at most equal to 1 and this happens exactly at the point [tex](x_1,x_2,x_3,\ldots)=(1,0,0,\ldots).[/tex]

The absolute minimum at the boundary of this function happens when all the variables are equal to 0 except [tex]x_n=\frac{1}{\sqrt{n}}[/tex] and this minimum is equal to c=1/n. To see this notice that

[tex]nf=nx_1^2+nx_2^2+\cdots nx_n^2\geq x_1^2+2x_2^2+\cdots nx_n^2=1[/tex]

(the equality sign is because now we are on the boundary). We notice that nf is greater than or equal to 1 and the minimum of nf=1 (this implies the minimum for f to be 1/n) is attained exactly when [tex](x_1,x_2,\ldots,x_n)=(0,0,\ldots,\frac{1}{\sqrt{n}})[/tex].

So, finally, [tex]a+b+c=0+1+\frac{1}{n}=\frac{n+1}{n}.[/tex]

Answer: A . A Sample

Step-by-step explanation:

In online class , there should be male students too.

Thus , the population : All students who take part in our online class

So the data of all female students who take part in our online class is just a sample of the entire population.

∴ All female students who take part in our online class can be described as a sample.

Hence, the correct answer is A. A Sample .

Answer:

a)

0.5

option A

b)

0.6

c)

0.1

Step-by-step explanation:

The event A and B are independent so

P(A∩B)=P(A)*P(B)

P(A∩B)=P(0.2)*P(0.5)=0.10

a)

We have to find P(B'|A')

P(B'|A')=P(B'∩A')/P(A')

P(A)=0.2

P(A')=Asian project is not successful=1-P(A)=1-0.2=0.8

P(B)=0.5

P(B')=Europe project is not successful=1-P(B)=1-0.5=0.5

P(B'∩A')=Europe and Asia both project are not successful=P(A')*P(B')=0.8*0.5=0.4

P(B'|A')=P(B'∩A')/P(A')=0.4/0.8=0.5

This can be done by another independence property for conditional probability

P(B|A)=P(B)

P(B'|A')=P(B')

P(B'|A')=0.5

b)

Probability of at least one of two  projects will be successful means that the probability of success of Asia project or probability of success of Europe project  or probability of success of Europe and Asian project which is P(AUB).

P(AUB)=P(A)+P(B)-P(A∩B)

P(AUB)=0.2+0.5-0.1

P(AUB)=0.6

c)

Probability of only Asian project is successful given that at least one of the two projects is successful means that probability of success of project Asia while the project Europe is not successful denoted as P((A∩B')/(A∪B))=?

P((A∩B')/(A∪B))=P((A∩B')∩(A∪B))/P(A∪B)

P((A∩B')∩(A∪B))=P(A∩B')*P(A∪B)

P(A∩B')=P(A)*P(B')=0.2*0.5=0.10

P((A∩B')∩(A∪B))=0.1*0.6=0.06

P((A∩B')/(A∪B))=0.06/0.6=0.1

Final answer:

The probability of European project being unsuccessful is 40%. The probability that at least one project will be successful is 60%. Given that at least one project is successful, the probability that only the Asian project is successful is approximately 16.7%.

Explanation:

The student is learning about independent and mutually exclusive events in probability.

The question involves calculating the probability of certain outcomes given two independent events, A and B, with known probabilities P(A) = 0.2 and P(B) = 0.5.

(a) Probability of Both Projects Being Unsuccessful

Since events A and B are independent, events A' (Asian project not successful) and B' (European project not successful) are also independent. Therefore, the correct reasoning is:

a. Since the events are independent, then A' and B' are independent, too.

The probability of both A' and B' occurring is P(A')P(B') = (1 - P(A))(1 - P(B)) = (1 - 0.2)(1 - 0.5) = 0.8  imes 0.5 = 0.4.

Thus, the probability that the European project is also not successful given the Asian project is not successful is 0.4 or 40%.

(b) Probability that At Least One Project Will Be Successful

To find the probability that at least one project will be successful, we need to calculate 1 - P(A' AND B').

This gives us 1 - P(A')P(B') = 1 - 0.4

= 0.6 or 60%.

(c) Probability that Only the Asian Project Is Successful Given At Least One Is Successful

First, we calculate the probability of only the Asian project being successful, which equals P(A)P(B') = 0.2 x 0.5 = 0.1 or 10%.

Next, we calculate the probability of at least one project being successful, which we found to be 60%.

Therefore, the conditional probability is P(A and not B) / P(at least one successful), which equals 0.1 / 0.6 ≈ 0.167 or 16.7%

Answer:

Step-by-step explanation:

consider B in matrix form

We have

[tex]\left[\begin{array}{ccc}1&1&1\\1&1&0\\0&1&1\end{array}\right][/tex]

Reduce this to row echelon form

by R1= R1-R3

we get

[tex]\left[\begin{array}{ccc}1&0&0\\1&1&0\\0&1&1\end{array}\right][/tex]

Now R2-R1 gives Identity matrix in row echelon form.  So rank =3 hence this is a basis for R cube

to find (1,2,3) as linear combination of B

Let a, b, and c be the scalars such that

a(1,1,1)+b(1,1,0)+c(0,1,1) = (1,2,3)

Equate corresponding terms as

a+b= 1:   a+b+c =2:  a+c =3

Solving b = -1, c = 1 and a = 2

(1,2,3) = 2(1,1,1)-1(1,1,0)+1(0,1,1)

The slope of the line is 0.993. A positive slope indicates a positive linear relationship between city mpg and highway mpg.

The slope of a regression line represents the change in the dependent variable (in this case, highway mpg) for a one-unit increase in the independent variable (city mpg).

In this equation, the slope is given by 0.993. This means that for every one-unit increase in city mpg, the highway mpg is expected to increase by 0.993 units.

The positive numerical value of the slope indicates that there is a positive linear relationship between city mpg and highway mpg. This means that as the city mpg increases, the highway mpg also tends to increase.

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The slope of the given regression line is 0.993. The slope of a line in a linear regression equation represents the rate of change. This means that for every one-unit increase in city gas mileage, the highway gas mileage increases by about 0.993 units, on average.

The slope of the given regression line is 0.993 The slope of a line in a linear regression equation represents the rate of change. With respect to this specific question, the slope of 0.993 represents that for each increase of 1 mile per gallon (mpg) in city gas mileage, we expect the highway gas mileage to increase by 0.993 mpg on average.

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Answer:

[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]

Step-by-step explanation:

For this case we have these two functions:

[tex] x+y^2 = 12[/tex]   (1)

[tex] x+y=0[/tex]   (2)

And as we can see we have the figure attached.

For this case we select the x axis in order to calculate the area.

If we solve y from equation (1) and (2) we got:

[tex] y = \pm \sqrt{12-x}[/tex]

[tex] y = -x[/tex]

Now we can solve for the intersection points:

[tex] \sqrt{12-x} = -\sqrt{12-x}[/tex]

[tex] 12-x = -12+x[/tex]

[tex] 2x=24 , x=12[/tex]

[tex] \sqrt{12-x} =-x[/tex]

[tex] 12-x = x^2[/tex]

[tex] x^2 +x -12=0 [/tex]

[tex] (x+4)*(x-3) =0[/tex]

And the solutions are [tex] x =-4, x=3[/tex]

So then we have in total 3 intersection point [tex] x=12, x=-4, x=3[/tex]

And we can find the area between the two curves separating the total area like this:

[tex] \int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx[/tex]

[tex] \int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx[/tex]

We can separate the integrals like this:

[tex] \int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx[/tex]

For this integral [tex] \int_{-4}^3 |\sqrt{12-x} dx [/tex] we can use the u substitution with [tex]u = 12-x[/tex] and after apply and solve the integral we got:

[tex] \int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}[/tex]

The other integral:

[tex] \int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}[/tex]

And for the other integral:

[tex]2\int_{3}^{12}\sqrt{12-x} dx[/tex]

We can use the same substitution [tex] u = 12-x[/tex] and after replace and solve the integral we got:

[tex]2\int_{3}^{12}\sqrt{12-x} dx =36[/tex]

So then the final area would be given adding the 3 results as following:

[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]

The task is to sketch the region enclosed by the parabola [tex]x + y^2 = 12[/tex]and the line [tex]x + y = 0,[/tex] and then find the area of that region. The area can be found by integrating with respect to y, using vertical slices through the region and the points of intersection as limits.

The student has asked to sketch the region enclosed by two equations and then find the area of that region. It looks like there may be some confusion with the equations provided, as they seem to contain typos. Assuming the correct equations are [tex]x + y2 = 12[/tex] and [tex]x + y = 0[/tex], we can proceed by first graphing these two curves.

After sketching, you would see that the curve [tex]x + y2 = 12[/tex] forms a parabola that opens to the right, and the line [tex]x + y = 0[/tex]is a straight line that has a negative slope and passes through the origin. The region enclosed by these two would resemble a 'cap' shape, with the point of intersection providing the limits for integration.

To find the area of this region, you could choose to integrate with respect to y since the equations can be easily solved for x, and the structure of the parabola suggests that vertical slices would be easier to work with. The integration limits would be the y-values where the curves intersect. After setting up the integral, you would calculate the definite integral to find the area.

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Answer:

Step-by-step explanation:

x² +  b²/4a² = -c / a + b²/4a²

x² + (b/2a)² = -c/a + (b/2a)²

(x + b/2a)² = -c/a + (b/2a)² =  -c / a + b²/4a² = (-4ac+ b²)/4a²

(x + b/2a)² =  (-4ac+ b²)/4a²

√{(x + b/2a)²} = √{(-4ac+ b²)/4a²}

x + b/2a = √(-4ac+ b²) / √(4a²) = √(-4ac+ b²) / 2a = √( b²-4ac) / 2a

x + b/2a  =  √( b²-4ac) / 2a

x + b/2a -b/2a  =  {√( b²-4ac) / 2a } -b/2a

x = -b/2a +   {√( b²-4ac) / 2a }

x = {-b±√( b²-4ac)}/2a

A quadratic equation is an equation of the sort; ax^2 + bx + c =0. It can be solved by the formula method.

A quadratic equation is an equation of the sort; ax^2 + bx + c =0. One of the ways of solving a quadratic equation is the formula method which is being derived here.

From the step shown in the image in the question;

Collecting like terms;

x² +  b²/4a² = -c / a + b²/4a²

x² + (b/2a)² = -c/a + (b/2a)²

We can now write;

(x + b/2a)² = -c/a + (b/2a)²

Hence;

(x + b/2a)² =  (-4ac+ b²)/4a²

Taking the square root of both sides and solving for x

x =-b±√( b²-4ac)/2a

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Answer: The examples that use the stratified method are: (1).  A questionnaire is created to gauge student opinion on a new university cafeteria. A sample of 40 eshmen, 50 sophomores, 60 juniors, and 50 seniors is selected to fill out the questionnaire. (2). A potato field is believed to be infected with a plant disease. The field is divided into 10 equal areas, and 25 potatoes are selected from each area to be tested for the disease.

Step-by-step explanation: Stratified sampling technique is a type of sampling where the population under study has a number of distinct categories or sub-groups in which it is divided into. These categories or sub-groups are called strata and are defined by certain characteristics related to the variable or particular finding under interest. The sampling frame can be organized into separate mutually exclusive strata and then each ‘stratum’ is being sampled as an independent sub-population out of which individual elements can be randomly selected. In this case, each unit in a stratum, that is, each element in a group has a chance of being selected. With stratified sampling, the best result occurs when elements within strata are internally homogenous.

Answer:

The required equation would be [tex]\frac{dT}{dt}=k|M-T|[/tex]

Where, k is constant of proportionality.

Step-by-step explanation:

Differential equation : is an equation that contains one or more functions and their derivatives.

Where, derivatives shows the rate of change of a variable with respect to other variable.

Given,

Change in temperature T of coffee with respect to t ( time ) is proportional to difference between the fixed temperature M and temperature of coffee,

i.e. [tex]\frac{dT}{dt}\propto |M - T|[/tex]    ( if M > T, Change is M - T if M < T then change is T - M)

[tex]\frac{dT}{dt}=k|M-T|[/tex]

Where,

k = constant of proportionality.

Answer:

[tex] n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154[/tex]

[tex] n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456[/tex]

For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:

(B) The sample size of A is less than the sample size of B.

Step-by-step explanation:

For this case we have the following data given:

[tex] \bar X_A= 45[/tex] represent the sample mean for A

[tex] s_A= 6.45[/tex] represent the sample deviation for A

[tex] ME_A = 1.02[/tex] represent the margin of error for A

[tex] \bar X_B= 43[/tex] represent the sample mean for B

[tex] s_B= 7.84[/tex] represent the sample deviation for B

[tex] ME_B= 0.72[/tex] represent the margin of error for B

And for this case we are assuming that we have the same confidence level of 95%

For this case we an use the fact that the sample deviation is an unbiased estimator for the population deviation [tex]\hat \sigma = \hat s[/tex] and we can use the following formula for the margin of error of the sample mean the following formula:

[tex] ME= z_{\alpha/2} \frac{\hat s}{\sqrt{n}}[/tex]

For this case the value of the significance is given by [tex] \alpha =1-0.95 =0.05[/tex] and the value for [tex]\alpha/2 =0.025[/tex] , so then the value for [tex] z_{\alpha/2}[/tex] represent a quantile of the normal standard distribution that accumulates 0.025 of the area on each tail of the normal standard distribution and for this case is [tex] z_{\alpha/2}=\pm 1.96[/tex].

So then since we have the value for z if we solve for n from the margin of error formula we got:

[tex] n = \frac{\hat s^2}{(\frac{ME}{z})^2}[/tex]

And for the case A we can find the sample size and we got:

[tex] n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154[/tex]

And for the case B we can find the sample size and we got:

[tex] n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456[/tex]

For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:

(B) The sample size of A is less than the sample size of B.

The sample size of B is larger than the sample size of A and this can be determined by using the formula of margin of error.

Given :

To determine the sample size for both cases A and B, the formula of Margin of Error can be used:

[tex]\rm ME =z_{\frac{\alpha }{2}} \dfrac{\hat{s}}{\sqrt{n} }[/tex]

[tex]\rm n =\left(\dfrac{\hat{s}}{\dfrac{ME}{z}}\right)^2[/tex]

Now, for case A:

[tex]\rm n_A =\left(\dfrac{6.45}{\dfrac{1.02}{1.96}}\right)^2[/tex]

[tex]\rm n_A\approx 154[/tex]

Now, for case B:

[tex]\rm n_B =\left(\dfrac{7.84}{\dfrac{0.72}{1.96}}\right)^2[/tex]

[tex]\rm n_B \approx 456[/tex]

So, the sample size of B is larger than the sample size of A.

Therefore, the correct option is B) The sample size of A is less than the sample size of B.

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The result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]

The given parameters are:

[tex]\int\int R(2x-y) dA[/tex]

x^2 + y^2 = 4

Lines x = 0 and y = x

By polar coordinates, we have:

x = rcost and y = rsint

dA = rdrdt

Substitute x = rcost and y = rsint in 2x - y

2x - y = 2rcost - rsint

So, the integral becomes

[tex]\int\int R(2x-y) dA = \int\limits^a_b \int\limits^a_b ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]

The lines x = 0 and y = x imply that the integral varies from 0 to 2 and π/2 to π/4.

So, we have:

[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]

Rewrite as:

[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2\cos (t) - \sin(t) )\ r^2drdt[/tex]

Split the integral

[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} ( 2\cos (t) - \sin(t) ) dt \int\limits^2_0 r^2dr[/tex]

Integrate

[tex]\int\int R(2x-y) dA = [2\sin (t) - \cos(t)]\limits^{\pi/4}_{\pi/2} * [\frac{r^3}{3}]\limits^2_0[/tex]

Expand

[tex]\int\int R(2x-y) dA = [2\sin (\pi/2) + \cos(\pi/2) - 2\sin (\pi/4) - \cos(\pi/4)] * [\frac{2^3 - 0^3}{3}][/tex]

Simplify the above expression

[tex]\int\int R(2x-y) dA = [2*1 + 0 - \sqrt 2 - \frac{\sqrt 2}{2}] * [\frac{8}{3}][/tex]

[tex]\int\int R(2x-y) dA = [\frac{4 - 3\sqrt 2}{2}] * [\frac{8}{3}][/tex]

Evaluate the product

[tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]

Hence, the result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]

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To evaluate the given double integral in polar coordinates, we first express the given region in terms of polar coordinates. Then, we rewrite the double integral using the polar coordinate expressions and find the limits of integration. Finally, we evaluate the integral using the given limits.

To evaluate the double integral ∫∫R(2x−y)dA in polar coordinates, we first need to express the given region R in terms of polar coordinates. The region R is enclosed by the curve x^2+y^2=4, the line x=0, and the line y=x. In polar coordinates, the curve x^2+y^2=4 becomes r^2=4, or r=2. The line x=0 becomes θ=90°, and the line y=x becomes θ=45°. So, the region R is bounded by θ=0° to θ=45° and r=0 to r=2.

Next, we need to express the differential area element dA in polar coordinates. In Cartesian coordinates, dA represents the area element dx dy. In polar coordinates, dA can be expressed as dA=r dr dθ.

Now, we can rewrite the given double integral as ∫∫R(2x−y)dA = ∫∫R(2rcos(θ)−rsin(θ))r dr dθ. Substituting the limits of integration, we have the final form of the double integral as:

∫[0 to 45°] ∫[0 to 2] (2r2cos(θ)−r2sin(θ)) r dr dθ.

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Answer:  0.15

Step-by-step explanation:

As per given , the probability that customers who bought a new vehicle bought an SUV : P(SUV) = 0.20

The probability that customer bought a vehicle that was an SUV and in black color : P(SUV and black)  =0.03

Now by suing conditional probability formula,

If we have given that a customer bought an SUV, then the probability that it was black will be :

[tex]\text{P(Black}|\text{SUV})=\dfrac{\text{P(SUV and Black)}}{\text{P(SUV)}}[/tex]

[tex]=\dfrac{0.03}{0.20}=\dfrac{3}{20}=0.15[/tex]

Hence, the required probability is 0.15.

The probability that a customer who bought an SUV also bought a black SUV is 0.006, or 0.6% (expressed as a percentage).

To find the probability that a customer who bought an SUV also bought a black SUV, you can use conditional probability.

Let's define the following events:

A: A customer bought an SUV.

B: A customer bought a black SUV.

You are given that P(B|A) is the probability that a customer who bought an SUV also bought a black SUV, which is 3% or 0.03.

You want to find P(B|A), the probability that a customer who bought an SUV also bought a black SUV. You can use the following formula for conditional probability:

P(B|A) = (P(A and B)) / P(A)

Here, P(A and B) is the probability that a customer bought both an SUV and a black SUV, and P(A) is the probability that a customer bought an SUV.

You know that P(B|A) = 0.03 and P(A) = 0.20.

Now, you need to find P(A and B), the probability that a customer bought both an SUV and a black SUV. You can rearrange the formula:

P(A and B) = P(B|A) * P(A)

P(A and B) = 0.03 * 0.20

P(A and B) = 0.006

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Answer:

Reason is Matrix Multiplication Technique

Step-by-step explanation:

Matrix Multiplication: In matrix multiplication each element of a row is multiplied with each element of a column. So, row with all its zero entries is multiplied with all of the columns and making corresponding entries as zeros as well.

Therefore, A with a row having all zero entries also produces a row with all zero entries during multiplication with any matrix B.

Answer:  a. 242.5 pounds

b.  236 pounds

c . 208 and 278 pounds

The results are unlikely to be representative of all players in that​ sport's league because players randomly selected from championship sports team not the whole league.

Step-by-step explanation:

Given : Listed below are the weights in pounds of 11 players randomly selected from the roster of a championship sports team.

278    303    186    292    276    205    208    236    278    198    208

Mean = [tex]\dfrac{\text{Sum of weights of all players}}{\text{Number of players}}[/tex]

[tex]=\dfrac{278+303+186+292+276+205+208+236+278+198+208}{11}\\\\=\dfrac{2668}{11}\approx242.5\text{ pounds}[/tex]

For median , first arrange weights in order

186, 198 , 205 , 208, 208 , 236 , 276 , 278 ,278, 292 , 303

Since , number of data values is 11 (odd)

So Median = Middlemost value = 236 pounds

Mode = Most repeated value= 208 and 278

The results are unlikely to be representative of all players in that​ sport's league because players randomly selected from championship sports team not the whole league.

The mean, median and mode values of the distribution are :

Given the data :

Arrange the data in ascending order :

The mean = ΣX/ n

The median ;

The mode :

Therefore ;

Mean = 242.5 pounds

Median = 236 pounds

Mode = 208 and 278

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In this data set, the mean number of sibling pairs is 2.38, the median is 2, and the first and third quartiles are 2 and 3, respectively. The sample standard deviation would require a more complex calculation involving the mean and variance.

To calculate the relevant statistics for this data set, we need to use the formulas associated with each statistic.

1. The mean is the average number of sibling pairs in these families, calculated as the sum of all responses divided by the total number of responses. In this case, the mean is (1*13 + 2*22 + 3*15 + 4*6 + 5*3 + 7*1) / 60 = 2.38.

2. The median is the middle value in the ordered data set. Here, since we have 60 responses, the median is the average of the 30th and 31st values, which both fall within the '2 siblings' category. So, the median is 2.

3. To calculate the sample standard deviation, we first find the variance (the average of the squared differences from the mean). Then standard deviation is the square root of the variance. The exact calculation is quite lengthy, so you might want to use a statistical calculator for this.

4. The first quartile (Q1) is the value that separates the first 25% of the data. Because 25% of 60 equals 15, Q1 is also 2.

5. The third quartile (Q3) is the value that separates the first 75% of the data. Since 75% of 60 is 45, Q3 falls within the '3 siblings' category, so Q3 is 3.

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The explicit solution to the initial-value problem is y = (-5x^2 + 2x^5 + 4x^2y^2 - 9)/(10(1 + 4y^2)).

To solve the given initial-value problem (1 + x^4)dy + x(1 + 4y^2)dx = 0, with y(1) = 0, the method of separation of variables is applied. Rearrange terms to isolate y and x:

(1 + x^4)dy = -x(1 + 4y^2)dx

Now, integrate both sides:

∫(1 + x^4)dy = -∫x(1 + 4y^2)dx

Integrating, we get:

y + (x^5)/5 = -(x^2)/2 - 2x^2y^2/2 + C

Solve for C using the initial condition y(1) = 0:

0 + 1/5 = -1/2 - 4/2 + C

C = 9/10

Substitute C back into the equation:

y + (x^5)/5 = -(x^2)/2 - 2x^2y^2/2 + 9/10

Now, simplify and solve for y:

y = (-5x^2 + 2x^5 + 4x^2y^2 - 9)/(10(1 + 4y^2))

This is the explicit solution to the initial-value problem. It is essential to note that the obtained solution is implicit and may not have a simple form due to the nature of the given differential equation.

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

[tex]u\neq 0_{R2}[/tex]      

[tex]v\neq 0_{R2}[/tex]

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to [tex]R^{2x2}[/tex] using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

[tex]A=\left[\begin{array}{cc}3&1&p&2\end{array}\right][/tex]

We used the first vector ''u'' as the first column of the matrix A

We used the  second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

[tex]Det(A)=6-p[/tex]

We need this determinant to be different to zero

[tex]6-p\neq 0[/tex]

[tex]p\neq 6[/tex]

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that [tex]p\neq 6[/tex]

We can write : p ∈ IR - {6}

Notice that is [tex]p=6[/tex] ⇒

[tex]u=(3,6)[/tex]

[tex]v=(1,2)[/tex]

If we write [tex]3v=3(1,2)=(3,6)=u[/tex] , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.

Answer: There are 1 and half table spoons of the supplement that the client will be taking each day.

Step-by-step explanation:

Since we have given that

Quantity of cup of a dietary supplement = [tex]\dfrac{1}{2}[/tex]

Number of times a day = 3

So, the number of tablespoons of the supplement that the client will be taking each day would be

[tex]3\times \dfrac{1}{2}\\\\=\dfrac{3}{2}\\\\=1\dfrac{1}{2}[/tex]

Hence, there are 1 and half table spoons of the supplement that the client will be taking each day.

Final answer:

To solve the given initial value problem, we'll use the method of integrating factors. After rearranging the equation to standard form, we'll find the integrating factor and proceed to solve for y.

Explanation:

To find the solution of the given initial value problem, we'll use the method of integrating factors. First, we'll rearrange the equation to the standard form: y' + (6/t)y = t - 1/t + 1. Now, let's identify the integrating factor, which is e^(∫6/t dt) = e^(6ln|t|) = t^6. Multiply both sides of the equation with t^6 to get t^7y' + 6t^5y = t^7 - t^6 + t^6. Now, we can integrate both sides and solve for y.

There is a chance to get one or more number of number of matches.

An arrangement of objects where the order in which the objects are selected does not matter.

Given,

Let n be a positive integer.

sample n numbers a1, a2,..., an from the set {1,...,n} uniformly at random, with replacement.

Where 1<j where ai = aj.

We need to find the expected total number of matches

expected total number of matches=Sum of sample/ Total number of samples

=1+2+3+4+...n/n

=n(n+1)/n

=n+1

Hence, there is a chance to get one or more number of number of matches.

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To determine the expected total number of matches when picking n numbers from the set {1,...,n} with replacement, we find that each pair of picks has a 1/n chance of a match. Multiplying this by the total number of distinct pairs gives us an expectation of (n-1)/2 matches.

To calculate the expected total number of matches when n numbers are sampled from the set {1,...,n} uniformly at random with replacement, we consider the probability of a match for each pair of picks (i,j) with i < j. Since each number has a 1/n chance of being picked, the probability that two picks are a match, P(match), is similarly 1/n. Assuming all pairs are independent, the expected number of matches for a specific pair is thus 1/n.

Because there are a total of n(n-1)/2 such pairs in our selection (since we can choose 2 out of n elements in a combinatorial fashion), the expected total number of matches E(total matches) is the sum of the expectations for all pairs. Therefore, E(total matches) = n(n-1)/2 * (1/n).

Upon simplifying the expression, we obtain E(total matches) = (n-1)/2. Thus, this is the expected total number of matches for the given scenario.

Answer:

Cost function C(x) where x is the cubic feet of water used.

[tex]C(x) = 75 + 0.03x[/tex]

Step-by-step explanation:

We are given the following in the question:

The municipal water supply are billed a fixed amount monthly plus a charge for each cubic foot of water used.

Let x be the fixed amount and y be the cost for each cubic foot of water used.

A household using 1700 cubic feet was billed $126.

Thus, we can write:

[tex]x + 1700y = 126[/tex]

A household using 2500 cubic feet was billed $150.

[tex]x + 2500y = 150[/tex]

Solving the two equation by eliminating x by subtracting the equations, we get,

[tex]x+2500y-(x + 1700y) = 150-126\\800y = 24\\\\y =\dfrac{24}{800} = 0.03\\\\x+1700(0.03) = 126\\\Rightarrow x = 126-(1700)(0.03) = 75[/tex]

Thus, the fixed cost is $75 and the cost per cubic foot of water is $0.03

Cost function can be written as:

[tex]C(x) = 75 + 0.03x[/tex]

where x is the cubic feet of water used.

Final answer:

To derive the total cost equation for the water bill, a system of linear equations was solved to find the fixed charge and the variable rate per cubic foot. The cost function is C(x) = 75 + 0.03x, where C represents the total cost and x represents cubic feet of water used.

Explanation:

To find the equation that represents the total cost of a resident's water bill as a function of the cubic feet of water used, we first need to establish the fixed charge and the variable charge per cubic foot. We do this by setting up a system of linear equations based on the information provided about the two households.

Let F be the fixed monthly charge and V be the variable charge per cubic foot. We can set up two equations based on the given information:

Subtracting the first equation from the second gives us:

800V = 24

Dividing both sides by 800, we find:

V = 0.03

Now, we plug the value of V back into the first equation to find F:

F + 1700(0.03) = 126

F + 51 = 126

F = 126 - 51

F = 75

Therefore, the equation for the total cost C, in dollars, as a function of x, the cubic feet of water used, is:

C(x) = 75 + 0.03x

Answer:

[tex]a'(t) = 2k*v'(t)*v(t)[/tex]

Step-by-step explanation:

According to the data provided, the acceleration can be modeled by the following equation:

[tex]a(t) = kv(t)^2[/tex]

Where a(t) is the acceleration as a function of time, and v(t) is the velocity ad a function of time.

Applying the chain rule, the differential equation, with proportionality constant k, is:

[tex]\frac{d(a(t))}{dt}=\frac{d(kv(t)*v(t))}{dt} \\a'(t) = k*(v(t)*v'(t)+v'(t)*v(t))\\a'(t) = 2k*v'(t)*v(t)[/tex]

To show AB and CD are independent events, we used the fact that A, B, C, and D are mutually independent. The calculations show that P(AB AND CD) equals P(AB)P(CD), satisfying the condition for independence of events AB and CD.

To demonstrate that events AB and CD are independent, we need to use the definition of mutual independence. By this definition, being mutually independent, events A, B, C, and D satisfy the condition that for any two distinct events, say A and B, P(A AND B) = P(A)P(B). Similarly, this extends to any three events and all four events together, giving us P(A AND B AND C) = P(A)P(B)P(C), and P(A AND B AND C AND D) = P(A)P(B)P(C)P(D).

Sine AB and CD are composed of mutually independent events, we can deduce the following:

Now, to show that AB and CD are independent, we need to verify if P(AB AND CD) = P(AB)P(CD).

Because A, B, C, and D are mutually independent, we can expand P(AB AND CD) as:

And since we already know the individual probabilities of AB and CD as shown above, we then have:

Observing that both P(AB AND CD) and P(AB)P(CD) result in the same product P(A)P(B)P(C)P(D), we conclude that AB and CD are indeed independent events.

The probability of seeing at least three spots when rolling a fair six-sided die is 2/3, using classical probability concepts. Each possible result, 3, 4, 5, or 6, is mutually exclusive and equally likely, so probabilities add to 2/3.

The subject of this question is probability, which is a branch of Mathematics. When a fair six-sided die is rolled, each face (1, 2, 3, 4, 5 and 6) has an equal probability of 1/6. If we want to find the probability that at least three spots will be showing up on the die, we are looking for the probability of getting a 3, 4, 5 or 6.

Since each possibility (getting a 3, 4, 5 or 6) is mutually exclusive and all are equally likely, we can just add these probabilities together. So, the probability of getting at least 3 spots on a roll is (1/6) + (1/6) + (1/6) + (1/6) = 4/6 = 2/3.

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Casey visited that loading dock 8 times.

In order to determine the number of times the forklift visited the loading dock, divide the total number of boxes by the total number of boxes the forklift can carry at once.

Number of visits = 23 / 3 = 7 2/3 = 8 times

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Final answer:

Casey had to visit the loading dock 8 times in total to move all 23 boxes, given that his forklift can carry 3 boxes per trip.

Explanation:

To calculate how many times Casey had to visit the loading dock to move 23 huge boxes with his forklift, which can only hold three boxes at once, we use division.

We divide the total number of boxes by the number of boxes the forklift can carry per trip. Since 23 divided by 3 gives us 7 with a remainder of 2, it means Casey made 7 full trips carrying 3 boxes each and one additional trip to carry the remaining 2 boxes.

Therefore, Casey had to visit the loading dock 8 times in total.

Answer:

b. False

Step-by-step explanation:

De Morgan's law states that considering two statements A and B;

                      not (A or B) = not A and not B; and                  

                       not (A and B) = not A or not B

In set theory;

                      [tex]\overline{A u B} = \overline{A} n \overline{B}\\\overline{A n B} = \overline{A} u \overline{B}[/tex]

Applying De Morgan's law to the question,

A = Kwame will take a job in industry

B = go to graduate school

  not (A or B) = Kwame will not take a job in industry and not go to graduate school

Also;  

  not (A and B) = Kwame will not take a job in industry or not go to graduate school

Now considering the question, answer provided "Kwame will not take a job in industry or will not go to graduate school." is FALSE