Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x + 6 = 0. (Round your answer to four decimal places.)

Answer:

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]

Step-by-step explanation:

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t}  )[/tex]

[tex]L(t^\frac{3}{2})=\int_{0 }^{\infty}t^\frac{3}{2}e^{-st}dt\\substitute \ u =st\\L(t^\frac{3}{2})=\int_{0 }^{\infty}\frac{u}{s} ^\frac{3}{2}e^{-u}\frac{du}{s}=\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}[/tex]

the integral is now in gamma function form

[tex]\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}=\frac{1}{s^{\frac{5}{2}}}\Gamma(\frac{5}{2})=\frac{1}{s^{\frac{5}{2}}}\times\frac{3}{2}\times\frac{1}{2} }\Gamma (\frac{1}{2} )=\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }[/tex]

now laplace of [tex]L(e^{-10t})[/tex]

[tex]L(e^{-10t})=\frac{1}{s+10}[/tex]

hence

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]

The Laplace transform of the function [tex]t^3/2 - e^-10t[/tex] is (3/2√π)/[tex]s^5/2 - 1/(s + 10)[/tex].

To find the Laplace transform of the function[tex]t^3/2 - e-10t[/tex], we use the Laplace transform properties and tables.

Answer: 0.75

Step-by-step explanation:

Given : Interval for uniform distribution : [0 minute, 5 minutes]

The probability density function will be :-

[tex]f(x)=\dfrac{1}{5-0}=\dfrac{1}{5}=0.2\ \ ,\ 0<x<5[/tex]

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-

[tex]P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75[/tex]

Hence,  the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75

The probability that a randomly selected passenger has a waiting time greater than 1.25 minutes is 1.

To find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes, we need to find the area under the probability density function (PDF) curve for values greater than 1.25. Since the waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes, the PDF is a rectangle with height 1/5 and base 5. The area of the rectangle represents the probability.

The probability of a waiting time greater than 1.25 minutes is the ratio of the area of the rectangle representing waiting times greater than 1.25 minutes to the total area of the rectangle representing all waiting times.

To calculate this probability, we first need to find the area of the rectangle representing waiting times greater than 1.25 minutes. Since the base of the rectangle is 5 minutes and the height is 1/5, the area is given by:

Area = base * height = 5 * (1/5) = 1

The total area of the rectangle representing all waiting times is the area of the entire rectangle, which is also equal to:

Area = base * height = 5 * (1/5) = 1

Therefore, the probability of a randomly selected passenger having a waiting time greater than 1.25 minutes is:

Probability = Area of waiting times greater than 1.25 minutes / Total area = 1/1 = 1

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Answer:

The probability of randomly selecting someone who does not believe that some people see the future in their dreams =0.497.

Step-by-step explanation:

Given

Percent of Americans who Say  they believe that some people see the future in their dreams=50.3%

Total percentages=100%

Therefore, Number of americans who say they believe that some people see the future in their dreams=50.3

The probability of randomly selecting someone  who say  they believe that some people see the future in their dreams =[tex]\frac{50.3}{100}[/tex]

Hence, the probability of randomly selecting someone who believe that some people see the future in their dreams, P(E)=0.503

Now, the probability of randomly selecting someone who does not believe that some people see the future in their dreams ,P(E')= 1-P(E)

The probability of randomly selecting someone who does not believe that some people see the future in their dreams =1-0.503

Hence,the probability of randomly selecting someone who does not believe that some people see the future in their dreams=0.497.

Answer: 0.497

Step-by-step explanation:

Let A be the event that Americans believe that some people see the future in their dreams.

Then , the probability that Americans believe that some people see the future in their dreams is given by :-

[tex]P(A)=50.3\%=0.503[/tex]

We know that the complement of a event X is given by :-

[tex]P(X')=1-P(X)[/tex]

Hence, the probability of randomly selecting someone who does not believe that some people see the future in their dreams is

[tex]P(A')=1-P(A)\\\\=1-0.503=0.497[/tex]

Answer:

Step-by-step explanation:

In order to find the equations we need the circle's general equation:

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

(h,k) is the center and 'r' is the radius.

A. Because the center is (-1,4) then h=-1 and k=4.

Now we can find the radius as:

[tex]distance=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}[/tex]

[tex]distance=\sqrt{(3-(-1))^{2}+(7-4)^{2}}[/tex]

[tex]distance=5[/tex] so we have r=5

Then the equation is [tex](x+1)^{2}+(y-4)^{2}=25[/tex]

B. Because we have two points defining a diameter we can find the radius as follows:

[tex]diameter=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}[/tex]

[tex]diameter=\sqrt{(7-(-3))^{2}+(13-(-11))^{2}}[/tex]

[tex]diameter=26[/tex]

[tex]radius=26/2=13[/tex]

Now let's find the center of the circle as follows:

[tex]C=(\frac{x1+x2}{2} , \frac{y1+y2}{2})[/tex]

[tex]C=(\frac{7-3}{2} , \frac{13-11}{2})[/tex]

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

Then the equation is [tex](x-2)^{2}+(y-1)^{2}=169[/tex]

Answer:

[tex]2\frac{11}{12}[/tex] cups of flour are nedeed

Step-by-step explanation:

we know that

3 dozen peanut butter cookies calls for 1 and 1/4 cups of flour

Convert mixed number to an improper fraction

[tex]1\frac{1}{4}\ cups=\frac{1*4+1}{4}=\frac{5}{4}\ cups[/tex]

using proportion

Find out how much flour would you need to make 7 dozen cookies

Let

x ----> the number of cups of flour

[tex]\frac{3}{(5/4)}\frac{dozen}{cups}=\frac{7}{x}\frac{dozen}{cups} \\ \\x=7*(5/4)/3\\ \\x=\frac{35}{12}\ cups[/tex]

Convert to mixed number

[tex]\frac{35}{12}\ cups=\frac{24}{12}+\frac{11}{12}=2\frac{11}{12}\ cups[/tex]

Answer:

2.92 or 2 and 23/25 cups are required for 7 dozen cookies

Step-by-step explanation:

Determine the required number of flour for 3 dozen cookies. Use it to find flour required for 7 dozen cookies.

Dozen cookies               Flour

        3                             1 + 1/4

        7                                x

Cross multiply to find the value of x

3x = 7(1+1/4)

3x = 7(5/4)

12x = 35

x = 2.92 cups or 2 and 23/25 cups

Therefore, 2.92 cups or 2 and 23/25 cups of flour are required for 7 dozen peanut butter cookies.

!!

[tex]t\dfrac{\mathrm dy}{\mathrm dt}-y^2\ln t+y=0[/tex]

Divide both sides by [tex]y(t)^2[/tex]:

[tex]ty^{-2}\dfrac{\mathrm dy}{\mathrm dt}-\ln t+y^{-1}=0[/tex]

Substitute [tex]v(t)=y(t)^{-1}[/tex], so that [tex]\dfrac{\mathrm dv}{\mathrm dt}=-y(t)^{-2}\dfrac{\mathrm dy}{\mathrm dt}[/tex].

[tex]-t\dfrac{\mathrm dv}{\mathrm dt}-\ln t+v=0[/tex]

[tex]t\dfrac{\mathrm dv}{\mathrm dt}-v=\ln t[/tex]

Divide both sides by [tex]t^2[/tex]:

[tex]\dfrac1t\dfrac{\mathrm dv}{\mathrm dt}-\dfrac1{t^2}v=\dfrac{\ln t}{t^2}[/tex]

The left side can be condensed as the derivative of a product:

[tex]\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac1tv\right]=\dfrac{\ln t}{t^2}[/tex]

Integrate both sides. The integral on the right side can be done by parts.

[tex]\displaystyle\int\frac{\ln t}{t^2}\,\mathrm dt=-\frac{\ln t}t+\int\frac{\mathrm dt}{t^2}=-\frac{\ln t}t-\frac1t+C[/tex]

[tex]\dfrac1tv=-\dfrac{\ln t}t-\dfrac1t+C[/tex]

[tex]v=-\ln t-1+Ct[/tex]

Now solve for [tex]y(t)[/tex].

[tex]y^{-1}=-\ln t-1+Ct[/tex]

[tex]\boxed{y(t)=\dfrac1{Ct-\ln t-1}}[/tex]

Answer:

14941.

Step-by-step explanation:

In base 16 we have that :

A=10, B=11, C=12, D=13, E=14, F=15 and the process of change is:

3: [tex]3*16^{0}= 3[/tex]

A5D= [tex]10*16^{2}+5*16^{1}+13*16^{0}= 2560+80+13=2653.[/tex]

3A5D =  [tex]3*16^{3}+10*16^{2}+5*16^{1}+13*16^{0}= 12288+2560+80+13=14941.[/tex]

The answer is:

[tex](f+g)(2)=20[/tex]

To solve the problem, we need to add the given functions, and then, evaluate the resultant function with the given value of "x" which is equal to 2.

We need to remember that:

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

So, we are given the functions:

[tex]f(x)=3x^2-2\\g(x)=4x+2\\[/tex]

Then, adding the functions , we have:

[tex](f+g)(x)=f(x)+g(x)=(3x^2-2)+(4x+2)[/tex]

[tex](f+g)(x)=3x^2-2+4x+2=3x^2+4x-2+2=3x^2+4x[/tex]

Therefore, we have that:

[tex](f+g)(x)=3x^2+4x[/tex]

Now, evaluating the function, we have:

[tex](f+g)(2)=3(2)^2+4(2)=3*4+4*2=12+8=20[/tex]

Hence, we have that the answer is:

[tex](f+g)(2)=20[/tex]

Have a nice day!

Answer:

15 subsets of cardinality 4 contain at least one odd number.

Step-by-step explanation:

Here the given set,

S={1,2,3,4,5,6},

Since, a set having cardinality 4 having 4 elements,

The number of odd digits = 3 ( 1, 3, 5 )

And, the number of even digits = 3 ( 2, 4, 6 )

Thus, the total possible arrangement of a set having 4 elements out of which atleast one odd number = [tex]^3C_1\times ^3C_3+^3C_2\times ^3C_2+^3C_3\times ^3C_1[/tex]

By using [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex],

[tex]=3\times 1+3\times 3+1\times 3[/tex]

[tex]=3+9+3[/tex]

[tex]=15[/tex]

Hence, 15 subsets of cardinality 4 contain at least one odd number.

Answer: a) The probability that a randomly selected high school student plays organized​ sports = 0.17

b) The randomly selected high school student is unlikely to play organized​ sports.

Step-by-step explanation:

Given : A survey of 400 randomly selected high school students determined that 68 play organized sports.

Then , the probability that a randomly selected high school student plays organized​ sports is given by :-

[tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{68}{400}=0.17[/tex]

In percent , the  probability that a randomly selected high school student plays organized​ sports is 17%.

since 17% lies in the interval of unlikely events (0%, 25%).

It means that the randomly selected high school student is unlikely to play organized​ sports.

The probability that a randomly selected high school student plays organized sports is 0.17, which means there is a 17% chance, or a 1 in 6 chance, that a randomly picked student plays organized sports.

This question relates to the field of Statistics, particularly to the concept of Probability. In this problem, we are provided with a total number of High School students (400), and a number of students who play organized sports (68).

To find the probability, you would divide the number of students who play sports by the total number of students. That is, Probability = Number with desired characteristic (play sports) / Total number In this case, it would be 68 / 400 = 0.17

Interpreting the probability means describing what it represents in practical terms. Here, a probability of 0.17 means that if you were to randomly select a high school student, there is a 17% chance, or around 1 in 6 chance, that this student plays organized sports.

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Answer:

Step-by-step explanation:

We are given that a and b are rational numbers where [tex]b\neq0[/tex] and x is irrational number .

We have to prove a+bx is irrational number by contradiction.

Supposition:let  a+bx is a rational number then it can be written in [tex]\frac{p}{q}[/tex] form

[tex]a+bx=\frac{p}{q}[/tex] where [tex]q\neq0[/tex] where p and q are integers.

Proof:[tex]a+bx=\frac{p}{q}[/tex]

After dividing p and q by common factor except 1 then we get

[tex]a+bx=\frac{r}{s}[/tex]

r and s are coprime therefore, there is no common factor of r and s except 1.

[tex]a+bx=\frac{r}{s}[/tex] where r and s are integers.

[tex]bx=\frac{r}{s}-a[/tex]

[tex]x=\frac{\frac{r}{s}-a}{b}[/tex]

When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.

Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.

Conclusion: a+bx is an irrational number.

[tex]X[/tex] has PDF

[tex]f_X(x)=\begin{cases}\frac1{10}&\text{for }0\le x\le10\\0&\text{otherwise}\end{cases}[/tex]

and thus CDF

[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\\frac x{10}&\text{for }0\le x\le10\\1&\text{for }x>10\end{cases}[/tex]

Because [tex]X[/tex] is continuous, we have

[tex]P(X<1\text{ or }X>8)=1-P(1\le X\le8)=1-(P(X\le8)-P(X\le1))[/tex]

[tex]P(X<1\text{ or }X>8)=1-P(X\le 8)+P(X\le1)[/tex]

[tex]P(X<1\text{ or }X>8)=1-F_X(8)+F_X(1)[/tex]

[tex]P(X<1\text{ or }X>8)=1-\dfrac8{10}+\dfrac1{10}=\boxed{\dfrac3{10}}[/tex]

Answer:

The entry on the second row and second column of the product matrix is [tex]c_{22} = 40[/tex].

Step-by-step explanation:

Let's define as A and B the given matrixes:

[tex]A = \left[\begin{array}{cc}1&2\\3&4\end{array}\right][/tex]

[tex]B = \left[\begin{array}{cc}9&6\\5&7\end{array}\right][/tex]

The product matrix C entry in the first row and first column [tex]c_{1,1}[/tex] or [tex]c_{11}[/tex] can be computer multiplying first row of A by first column of B (see example attached).

The product matrix C entry in the first row and second column [tex]c_{1,2}[/tex] or [tex]c_{12}[/tex] can be computer multiplying first row of A by second column of B.

The product matrix C entry in the second row and first column [tex]c_{2,1}[/tex] or [tex]c_{21}[/tex] can be computer multiplying second row of A by first column of B.

The product matrix C entry in the second row and second column [tex]c_{2,2}[/tex] or [tex]c_{22}[/tex] can be computer multiplying second row of A by second column of B.

Then, let's compute [tex]c_{22}[/tex] by doing the dot product between [3 4] and [6 7]...

[tex]c_{22} = [3 4] . [6 7] = 3*4 + 4*7 = 12 + 28 = 40[/tex]

Answer:

3.8%

Step-by-step explanation:

Simple interest formula is I=P*R*T

where:

I=interest

P=principal

R=rate

T=time.

So let's plug in our information we are given:

I=855

P=3750

T=6

R=?.

The equation becomes 855=3750*R*6.

Multiplication is commutative so we could write this as 3750*6*R=855.

After the multiplication of 3750 and 6 we obtain 22500*R=855.

Now we just divide both sides by 22500.  This will give us:

R=855/22500 which when entered into the calculator as 855 division sign 22500 gives us 0.038 or 3.8%.

We can use reduction of order. Given that [tex]y_1(x)=x[/tex] is a known solution, we look for a solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex]. It has derivatives [tex]{y_2}'=v'y_1+v{y_1}'[/tex] and [tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]. Substituting these into the ODE gives

[tex]x(xv''+2v')-x(xv'+v)+xv=0[/tex]

[tex]x^2v''+(2x-x^2)v'=0[/tex]

Let [tex]w(x)=v'(x)[/tex] so that [tex]w'(x)=v''(x)[/tex] and we get an ODE linear in [tex]w[/tex]:

[tex]x^2w'+(2x-x^2)w=0[/tex]

Divide both sides by [tex]e^x[/tex]:

[tex]x^2e^{-x}w'+(2x-x^2)e^{-x}w=0/tex]

Since [tex](x^2e^{-x})=(2x-x^2)e^{-x}[/tex], we can condense the left side as the derivative of a product:

[tex](x^2e^{-x}w)'=0[/tex]

Integrate both sides and solve for [tex]w(x)[/tex]:

[tex]x^2e^{-x}w=C\implies w=\dfrac{Ce^x}{x^2}[/tex]

Integrate both sides again to solve for [tex]v(x)[/tex]. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:

[tex]v=\displaystyle C_2+C_1\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

where [tex]x_0[/tex] is any point on an interval over which a solution to the ODE exists.

Finally, multiply by [tex]y_1(x)[/tex] to solve for [tex]y_2(x)[/tex]:

[tex]y_2=\displaystyle C_2x+C_1x\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

[tex]y_1(x)[/tex] already accounts for the [tex]C_2x[/tex] term above, so the second independent solution is

[tex]y_2=x\displaystyle\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

Answer:

future payment is $77056.92

total interest is paid after 45 year is  $23056.42

Step-by-step explanation:

Given data

payment (P) = $100

No of installment (n) = 12

rate of interest ( r ) = 1.5 %  i.e. = 0.015

time period (t) = 45 years

to find out

future payment and interest after 45 year

solution

we know future payment formula i.e. given below

future payment = payment × [tex](1+\frac{r}{n})^{nt} - 1) / (r/n)[/tex]

now put all these value in equation

future payment = $ 100  × [tex](1+\frac{0.015}{12})^{12*45} - 1) / (0.015/12)[/tex]  

future payment = $ 77056.92

payment paid in 45 year @ $100 total money is paid is 45 × 12 × $100 i.e. = $54000

total interest = future payment  - money paid

total interest = $77056.42 - $54000

total interest = $23056.42

P - existing home sales will decrease

Q - new housing starts will increase

R - unemployment will decrease

S - the Federal Reserve decreases long-term interest rates

T - foreign trade deficits decrease

U - foreign trade deficits increase

V - manufacturing rates increase

[tex][(P \vee Q) \wedge R] \iff (S \wedge T) \vee (U \wedge V)[/tex]

P - Germany will vote to limit the number of immigrants it admits

Q - will reject Angela Merkel’s international policies

R - neighboring EU countries agree to a multi-national work visa program

S - the World Bank revalues the Euro relative to the US dollar

T - the US-Russia brokered peace treaty is signed by Syria.

[tex]\neg (R \wedge (S \vee T))\implies (P \implies R)\\[/tex]

Answer:

The system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.

Step-by-step explanation:

The given educations are

[tex]x+2y+z=-4[/tex]

[tex]-2x-3y-z=2[/tex]

[tex]2x+4y+2z=-8[/tex]

Using the Gauss-Jordan elimination method, we get

[tex]\begin{bmatrix}1 & 2 & 1\\ -2 & -3 & -1\\ 2 & 4 & 2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ -8\end{bmatrix}[/tex]

[tex]R_3\rightarrow R_3-2R_1[/tex]

[tex]\begin{bmatrix}1 & 2 &1\\ -2 & -3 & -1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ 0\end{bmatrix}[/tex]

Since elements of bottom row are 0, therefore the system of equations have infinitely many solutions.

[tex]0x+0y+0z=0\Rightarrow 0=0[/tex]

[tex]R_2\rightarrow R_2+2R_1[/tex]

[tex]\begin{bmatrix}1 & 2 &1\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ -6\\ 0\end{bmatrix}[/tex]

[tex]R_1\rightarrow R_1-R_2[/tex]

[tex]\begin{bmatrix}1 & 1 &0\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}2\\ -6\\ 0\end{bmatrix}[/tex]

[tex]x+y=2[/tex]

[tex]y+z=-6[/tex]

Let x=t

[tex]t+y=2\rightarrow y=2-t[/tex]

The value of y is 2-t.

[tex](2-t)+z=-6[/tex]

[tex]z=-6-2+t[/tex]

[tex]z=t-8[/tex]

The value of z is t-8.

Therefore the he system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.

To solve the system of linear equations using the Gauss-Jordan elimination method, perform row operations on the augmented matrix to obtain the reduced row-echelon form. The solution is x = -2, y = 1, and z = 0.

To solve the system of linear equations using the Gauss-Jordan elimination method, we need to perform row operations to transform the augmented matrix into row-echelon form and then into reduced row-echelon form. Let's start by representing the system of equations as an augmented matrix:

[1 2 1 -4; -2 -3 -1 2; 2 4 2 -8]

Performing row operations, you can transform the augmented matrix into reduced row-echelon form, obtaining:
[1 0 0 -2; 0 1 0 1; 0 0 1 0]

The solution to the system is x = -2, y = 1, and z = 0. Therefore, (x, y, z) = (-2, 1, 0).

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To find the probability of picking an apple with a diameter greater than 3.76 inches from a normally distributed population with a mean of 3 inches and a standard deviation of 0.5 inches, we calculate a z-score and then find the corresponding probability using a z-table or calculator.

The question is about finding the probability in a situation that follows a normal distribution, specifically, the likelihood of picking an apple with a diameter greater than 3.76 inches given a mean of 3 inches and a standard deviation of 0.5 inches.

To solve this, we must first calculate the z-score, which is the number of standard deviations a data point (in this case, the apple size of 3.76 inches) is from the mean. The z-score is calculated as follows: Z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation. Substituting the given values, we get Z = (3.76 - 3) / 0.5, which results in a z-score of 1.52.

Next, we will refer to the z-table to find the probability corresponding to this z-score. However, most z-tables give the probability from the mean to our z-score (the left-hand side). But we want the probability that the apple's diameter is greater than 3.76 inches, i.e., the right-hand side of the distribution curve. As a result, we have to subtract the value we get from the z-table from 1 (because the cumulative probability under the entire normal curve is 1).

If a z-table is not readily available, there are many online calculators available to find this probability.

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Answer:

The original price is = $75

The reduced price = $50

So, price reduced is = [tex]75-50=25[/tex] dollars

Total swimsuits are = [tex]40+25=65[/tex]

Total markdown = [tex]65\times25=1625[/tex] dollars

Now, 25 swimsuits were returned to the original price. Means 25 swimsuits were returned to $75, increasing $25 again.

So, markdown cancellation = [tex]25\times25=625[/tex] dollars

Net markdown = total markdown - markdown cancellation

= [tex]1625-625=1000[/tex] dollars

Answer:

  215

Step-by-step explanation:

The 238 taking Finite Math includes those taking Finite Math and Statistics. Subtracting out the 23 who are taking both leaves 215 taking Finite Math only.

215 students are taking Finite Mathematics but not Statistics.

To find out how many students are taking Finite Mathematics but not Statistics at Southern States University (SSU), let's break down the information given and use set theory concepts.

Total number of students taking either Finite Mathematics or Statistics: 399

Number of students taking Finite Mathematics: 238

Number of students taking Statistics: 184

Number of students taking both Finite Mathematics and Statistics: 23

First, we need to figure out how many students are taking only Finite Mathematics. We can do this by subtracting the number of students taking both Finite Mathematics and Statistics from the total number of students taking Finite Mathematics.

Number of students taking only Finite Mathematics = Total taking Finite Mathematics - Total taking both Finite Mathematics and Statistics

So,

Number of students taking only Finite Mathematics = 238 - 23

Number of students taking only Finite Mathematics = 215

Set up a ratio:

You drove 72 minutes and 100 km = 72/100

You want the number of minutes (x) to drive 150 km = x/150

Set the ratios to equal each other and solve for x:

72/100 = x/150

Cross multiply:

(72 * 150) = 100 * x)

Simplify:

10,800/100x

Divide both sides by 100:

x = 10800/100 = 108

This means it would take 108 minutes to drive 150 km.

Now subtract the time you have already driven to fin how much more you need:

180 - 72 = 36 more minutes.

Answer:

36 min

Step-by-step explanation:

It takes 72 min to drive 100 km. 50 km are left to drive.

Half of the driving above is: It takes 36 min to drive 50 km.

[tex]\displaystyle\pi\int_2^5\left(6^2-\left(\frac43x-\frac23\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(20+x-x^2)\,\mathrm dx=\boxed{56\pi}[/tex]

[tex]\displaystyle2\pi\int_2^5x\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5x(5-x)\,\mathrm dx=\boxed{36\pi}[/tex]

[tex]\displaystyle2\pi\int_2^5(7-x)\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5(35-12x+x^2)\,\mathrm dx=\boxed{48\pi}[/tex]

[tex]\displaystyle\pi\int_2^5\left((6-2)^2-\left(\frac43x-\frac23-2\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(5+4x-x^2)\,\mathrm dx=\boxed{32\pi}[/tex]

The y-coordinates of the two intersection points of the triangle and the line y=2 (Option d).

In this explanation, we will explore how to find the volumes of solids formed by revolving a triangle with given vertices about different axes and lines. We'll use basic calculus principles to calculate the volumes and understand the concept of rotation in three-dimensional space.

a) To find the volume of the solid generated by revolving the triangle about the x-axis, we imagine rotating the triangle in a circular motion around the x-axis. This forms a three-dimensional shape known as a "solid of revolution."* To calculate the volume, we integrate the cross-sectional area of each infinitesimally thin slice of the solid perpendicular to the x-axis, from the x-coordinate of the leftmost point to the rightmost point.

Let's use the "disk method" to integrate the cross-sectional areas. Each disk has a radius equal to the y-coordinate of the triangle at a particular x-coordinate. The formula for the volume using the disk method is:

Vx = ∫[from a to b] π * (y)² dx

Where (a, b) are the x-coordinates of the leftmost and rightmost points of the triangle, and y represents the y-coordinate of the triangle at a specific x.

b) Similarly, to find the volume of the solid formed by revolving the triangle about the y-axis, we use the "washer method". In this case, the inner radius of each washer is given by the x-coordinate of the triangle at a particular y-coordinate. The formula for the volume using the washer method is:

Vy = ∫[from c to d] π * (x)² dy

Where (c, d) are the y-coordinates of the bottommost and topmost points of the triangle, and x represents the x-coordinate of the triangle at a specific y.

c) To find the volume of the solid formed by revolving the triangle about the line x=7, we use the "shell method".

We integrate the circumference of each cylindrical shell formed between the triangle and the line x=7. The formula for the volume using the shell method is:

V7 = ∫[from e to f] 2π * (x-7) * y dx

Where (e, f) are the x-coordinates of the two intersection points of the triangle and the line x=7.

d) Lastly, to find the volume of the solid formed by revolving the triangle about the line y=2, we can use the shell method as well, considering cylindrical shells formed between the triangle and the line y=2. The formula for the volume using the shell method is:

Vy=2 = ∫[from g to h] 2π * (y-2) * x dy

Where (g, h) are the y-coordinates of the two intersection points of the triangle and the line y=2.

By calculating the integrals using these formulas, we can find the volumes of the solids generated by revolving the triangle about the specified axes and lines. Remember to always set up the integral limits correctly based on the x or y coordinates of the triangle's vertices.

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Answer:

a. 37 rooms are need to break even.

b. 73 rooms are required to make twice as much profit as you spent.

Step-by-step explanation:

Revenue of the company from each additional room is $1200.

Total construction cost = $44,000.

At break even condition, total revenue is equal to total cost. In other words, the profit of the firm is zero at break even.

Let x be the number of rooms that are need to break even.

Total revenue of x room is

[tex]TR=1200x[/tex]

At break even,

Total revenue = Total cost

[tex]1200x=44000[/tex]

Divide both sides by 1200.

[tex]x=\frac{44000}{1200}=36.6667\approx 37[/tex]

Therefore, 37 rooms are need to break even.

Let y be the number of rooms to make twice as much profit as you spent.

Total revenue of y room is

[tex]TR=1200y[/tex]

Total revenue = 2 × Total cost

[tex]1200x=2\times 44000[/tex]

[tex]1200x=88000[/tex]

Divide both sides by 1200.

[tex]y=\frac{88000}{1200}=73.33\approx 73[/tex]

Therefore 73 rooms are required to make twice as much profit as you spent.

Answer:

0.592

Step-by-step explanation:

The volume of the solid, rounded to three decimal places is 18.257.

The volume of the solid obtained by rotating the region around the y-axis can be found using the method of discs or washers. Since the region is bounded by the x-axis, the y-axis, the line y=2, and the curve [tex]y=e^x[/tex], we will integrate with respect to y.

The volume V of the solid of revolution is given by the integral:

[tex]\[ V = \pi \int_{a}^{b} [R(y)]^2 dy - \pi \int_{a}^{b} [r(y)]^2 dy \][/tex]

where [tex]\( R(y) \)[/tex] is the outer radius and [tex]\( r(y) \)[/tex] is the inner radius of the discs or washers.

In this case, the outer radius [tex]\( R(y) \)[/tex] is given by the line y=2, which is a horizontal line, so the outer radius is constant and equal to 2. The inner radius [tex]\( r(y) \)[/tex] is given by the curve [tex]y=e^x[/tex]. To express x in terms of y, we take the natural logarithm of both sides to get [tex]\( x = \ln(y) \)[/tex].

Now we can set up our integrals:

[tex]\[ V = \pi \int_{0}^{2} [2]^2 dy - \pi \int_{0}^{2} [\ln(y)]^2 dy \][/tex]

[tex]\[ V = \pi \int_{0}^{2} 4 dy - \pi \int_{0}^{2} [\ln(y)]^2 dy \][/tex]

The first integral is straight forward:

[tex]\[ \pi \int_{0}^{2} 4 dy = \pi \left[ 4y \right]_{0}^{2} = \pi [4(2) - 4(0)] = 8\pi \][/tex]

The second integral requires integration by parts. Let [tex]\( u = [\ln(y)]^2 \)[/tex]and [tex]\( dv = dy \)[/tex], then [tex]\( du = \frac{2\ln(y)}{y} dy \)[/tex] and [tex]\( v = y \)[/tex]. Applying integration by parts gives:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ y[\ln(y)]^2 - \int_{0}^{2} 2\ln(y) dy \right] \][/tex]

Now, we need to integrate [tex]\( 2\ln(y) \)[/tex] by parts again, with [tex]\( u = \ln(y) \)[/tex] and [tex]\( dv = dy \)[/tex], then [tex]\( du = \frac{1}{y} dy \)[/tex] and [tex]\( v = y \)[/tex]. This gives:

[tex]\[ \int_{0}^{2} 2\ln(y) dy = \left[ 2y\ln(y) - \int_{0}^{2} 2 dy \right] \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = \left[ 2y\ln(y) - 2y \right]_{0}^{2} \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = 2(2)\ln(2) - 2(2) - (0) \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = 4\ln(2) - 4 \][/tex]

Putting it all together:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ y[\ln(y)]^2 - (4\ln(2) - 4) \right]_{0}^{2} \][/tex]

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ 2[\ln(2)]^2 - (4\ln(2) - 4) \right] - \pi \left[ \lim_{y \to 0} y[\ln(y)]^2 - (4\ln(2) - 4) \right] \][/tex]

The limit as y approaches 0 of [tex]\( y[\ln(y)]^2 \)[/tex] is 0, so we have:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ 2[\ln(2)]^2 - 4\ln(2) + 4 \right] \][/tex]

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = 2\pi[\ln(2)]^2 - 4\pi\ln(2) + 4\pi \][/tex]

Now, subtract this from the first integral:

[tex]\[ V = 8\pi - (2\pi[\ln(2)]^2 - 4\pi\ln(2) + 4\pi) \][/tex]

[tex]\[ V = 8\pi - 2\pi[\ln(2)]^2 + 4\pi\ln(2) - 4\pi \][/tex]

[tex]\[ V = 4\pi + 4\pi\ln(2) - 2\pi[\ln(2)]^2 \][/tex]

[tex]\[ V = 4\pi(1 + \ln(2) - \frac{1}{2}[\ln(2)]^2) \][/tex]

Rounded to three decimal places, the volume is:

[tex]\[ V \approx 4\pi(1 + \ln(2) - \frac{1}{2}[\ln(2)]^2) \approx 4\pi(1 + 0.693 - \frac{1}{2}(0.693)^2) \][/tex]

[tex]\[ V \approx 4\pi(1 + 0.693 - 0.240) \][/tex]

[tex]\[ V \approx 4\pi(1.453) \][/tex]

[tex]\[ V \approx 5.812\pi \][/tex]

[tex]\[ V \approx 18.257 \][/tex]

Therefore, the volume of the solid, rounded to three decimal places, is:

[tex]\[ \boxed{18.257} \][/tex].

The complete question is:

Find the volume of the solid obtained by rotating the region bounded by the x-axis, the y-axis, the line y=2, and [tex]y=e^x[/tex] about the y-axis. Round your answer to three decimal places.

Answer:

15 samples

Step-by-step explanation:

The total sample space consists of 6 items

{79,64,84,82,92,77}

So,

n=6

The instructor has to randomly select 2 test scores out of 6.

So, r=6

The arrangement of scores selection doesn't matter so combinations will be used.

[tex]C(n,r)=\frac{n!}{r!(n-r)!} \\C(6,2)=\frac{6!}{2!(6-2)!}\\=\frac{6!}{2!*4!}\\=\frac{6*5*4!}{2!*4!} \\=\frac{30}{2}\\=15\ ways[/tex]

Therefore, there are 15 different samples are possible without replacement ..

Answer:

Slope is 0.094,

It represents the average rate of change.

Step-by-step explanation:

Since, the slope is the ratio of difference in y-coordinates and the difference in x-coordinates,

Also, in a order pair, first element shows the x-coordinate and second element shows the y-coordinate.

Here, the data points are (1950, 0.75) and (1997, 5.15),

Thus, the slope is,

[tex]m=\frac{5.15-0.75}{1997-1950}[/tex]

[tex]=\frac{4.4}{47}[/tex]

[tex]=0.0936170212766[/tex]

[tex]\approx 0.094[/tex]

Also, Slope represents the average rate of change.

Answer: 0.3446

Step-by-step explanation:

Given  : Mean : [tex]\mu = 90[/tex]

Standard deviation : [tex]\sigma = 10[/tex]

Also, the reading speed of second grade students in a large city is approximately​ normal.

Then , the formula to calculate the z-score is given by :_

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x =  94

[tex]z=\dfrac{94-90}{10}=0.4[/tex]

The p-value = [tex]P(z>0.4)=1-P(z<0.4)=1-0.6554217[/tex]

[tex]\\\\=0.3445783\approx0.3446[/tex]

Hence, the  probability a randomly selected student in the city will read more than 94 words per​ minute =0.3446

Answer:20.91

Step-by-step explanation:

Given

Principal amount invested=[tex]\$ 1,000,000[/tex]

Rate of interest=8%

Annual Withdrawl=[tex]\$ 100,000[/tex]

compound interest is given by

A=[tex]\left (1+ \frac{r}{100}\right )^t[/tex]

Therefore reamining Amount after certain years

Net money will become zero after t year

[tex]1,000,000\left (1+ \frac{8}{100} \right )^t - 100,000\left ( \frac{\left ( 1.08\right )^{t}-1}{0.08}\right )[/tex]=0

[tex]0.8\left ( 1.08\right )^t=\left ( 1.08\right )^{t}-1[/tex]

t=20.91 years

4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.

Start with

[tex]x=7\cdot9+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 4, the last two terms vanish and we're left with

[tex]x\equiv63\equiv64-1\equiv-1\equiv3\pmod4[/tex]

We have [tex]3^2\equiv9\equiv1\pmod4[/tex], so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.

[tex]x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 7, the first and last terms vanish and we're left with

[tex]x\equiv72\equiv2\pmod7[/tex]

which is what we want, so no adjustments needed here.

[tex]x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 9, the first two terms vanish and we're left with

[tex]x\equiv140\equiv5\pmod9[/tex]

so we don't need to make any adjustments here, and we end up with [tex]x=401[/tex].

By the Chinese remainder theorem, we find that any [tex]x[/tex] such that

[tex]x\equiv401\pmod{4\cdot7\cdot9}\implies x\equiv149\pmod{252}[/tex]

is a solution to this system, i.e. [tex]x=149+252n[/tex] for any integer [tex]n[/tex], the smallest and positive of which is 149.

The problem is about finding a number x that satisfies a system of modular arithmetic equations. It can be solved using the Chinese Remainder Theorem which is part of number theory in mathematics. More information is needed to solve this specific system.

The problem at hand is to find a number x which satisfies the conditions x ≡ 1 (mod 4), x ≡ 2 (mod 7), and x ≡ 5 (mod 9). This falls under the mathematical concept of modular arithmetic.

Modular arithmetic is a system of arithmetic for integers, where numbers wrap around once reaching a certain value—the modulus.

The expressions x ≡ 1 (mod 4), x ≡ 2 (mod 7), and x ≡ 5 (mod 9) mean that when x is divided by 4, the remainder is 1; when x is divided by 7, the remainder is 2; and when x is divided by 9, the remainder is 5 respectively.

This is a type of problem known as a system of linear congruences, which can be solved by applying the Chinese Remainder Theorem. However, the information provided is insufficient to provide a specific numerical solution to the system of congruences. It is recommended that the student consults the section of their classroom material that discusses the Chinese Remainder Theorem and its applications.

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