Shawn is interested in purchasing a new computer system for $1,650.00 and would like to apply a down payment of 20%. Calculate the down payment amount. Round dollars to the nearest cent.

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:

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: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

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%.

Answer:

Matrix A is not invertible.

Step-by-step explanation:

Invertible matrix are those matrix which are square matrix or non singular matrix.

For any matrix to be invertible, matrix should be non singular i.e. det(x)[tex]\neq[/tex] 0.

But for the question given above we cannot find determinant of matrix A as it is not square matrix. so inverse of given matrix does not exist. so it is not possible to have non trival solutions.

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]

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.

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:

  a.  x'(t) = kx(5000-x)

  b.  R_in = 7 lbs/min

     R_out = 5c(t) lbs/min

     c(t) = A(t)/(800 -3t) lbs/gal

     A'(t) = R_in - R_out; A(0) = 200 lbs

Step-by-step explanation:

A. When proportionality is joint between x and y, the expression describing that is kxy, where k is the constant of proportionality. Here, the rate of change is jointly proportional to number who have adopted (x) and number of the community of 5000 who have not (5000-x). The corresponding equation is ...

  x'(t) = kx·(5000-x)

We know the initial number of adopters must be greater than 0, or the equation's only solution would be x=0.

__

B. The rate of influx of salt is ...

  R_in = (3.5 lbs/gal)×(2 gal/min) = 7 lbs/min

  The number of gallons in the tank at time t is ...

  v = 800 gal + (2 gal/min - 5 gal/min)(t min) = (800 -3t) gal

For amount of salt A(t) pounds, the concentration c(t) will be ...

  c(t) = A(t)/v = A(t)/(800 -3t) . . . . lbs/gal

The outflow rate will be the product of outflow volume and concentration:

  R_out = (c(t) lbs/gal)(5 gal/min) = 5c(t) lbs/min

And the differential equation for A(t) is ...

  A'(t) = R_in -R_out . . . with initial condition A(0) = 200 lbs.

__

When you go to solve the equation for A'(t), it will need to be cast in terms of A(t):

  A'(t) = 7 -5A(t)/(800 -3t)

Answer:

E = 3P + 3S,

S=(1/3)E-P

Step-by-step explanation:

Given,

A slice of pizza and a can of soda cost $3,

That is, total cost of P slices of pizza and S cans of soda = ( 3P + 3S ) dollars,

According to the question,

Total cost of P slices of pizza and S cans of soda = E dollars,

⇒ 3P + 3S = E

Which is the required equation that correctly describes the amount of money spent,

For rearranging the equation in terms of E and P,

We need to isolate S,

Subtract 3P on both sides,

3S = E - 3P

Divide both sides by 3,

[tex]S=\frac{E-3P}{3}[/tex]

[tex]\implies S=(\frac{1}{3})E-P[/tex]

Which is the required equation.

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.

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: (16.6%, 21.4%)

Step-by-step explanation:

The confidence interval for proportion is given by :-

[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]

Given : Sample size : n= 708

The proportion of respondents who had​ children = [tex]p=0.19[/tex]

Significance level : [tex]\alpha=1-0.90=0.1[/tex]

Critical value : [tex]z_{\alpha/2}=z_{0.05}=\pm1.645[/tex]

Now, the 90​% confidence interval for proportion will be :-

[tex]0.19\pm (1.645)\sqrt{\dfrac{0.19(1-0.19)}{708}}\approx0.19\pm 0.024\\\\=(0.19-0.024,0.19+0.024)=(0.166,\ 0.214)=(16.6\%,\ 21.4\%)[/tex]

Hence, the 90​% confidence interval for the proportion = (16.6%, 21.4%)

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:

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

[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.

To know more about Volumes here

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Answer:

a) 8259888

b) 34220

c) 45057474

Step-by-step explanation:

Given,

The total number of transistor = 65,

In which, the defective transistor = 6,

So, the number of non defective transistor = 65 - 6 = 59,

Since, out of these transistor 5 are selected,

a) Thus, the number of ways = the total possible combination of 5 transistors = [tex]{65}C_ 5[/tex]

[tex]=\frac{65!}{(65-5)!5!}[/tex]

[tex]=8259888[/tex]

b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = [tex]{6}C_3\times {59}C_2[/tex]

[tex]=\frac{6!}{(6-3)!3!}\times \frac{59!}{(59-2)!\times 2!}[/tex]

[tex]=20\times 1711[/tex]

[tex]=34220[/tex]

c) The number of sample without any defective transistor = The possible combination of 0 defective transistor = [tex]^6C_0\times ^{59}C_5[/tex]

[tex]=1\times 45057474[/tex]

[tex]=45057474[/tex]

(a) 8259888 ways, (b) 34220 samples and (c) 5006386 samples.

Let's solve each part:

(a) There are 65 total transistors, and we want to select 5 of them. So, the number of ways to select a sample is the number of combinations of 65 things taken 5 at a time. This can be calculated using the combination formula:

Number of combinations = [tex]C(n, k) = \frac{n!}{ k! \times (n-k)!)}[/tex]

where:

n is the total number of items (65 transistors)

k is the number of items to select (5 transistors)

Plugging in the values:

[tex]C(65, 5) = \frac{65!}{ 5! \times (65-5)!)} = 8259888[/tex]

(b) We want samples that contain exactly 3 defective transistors. We can achieve this by selecting 3 defective transistors from the 6 available and 2 non-defective transistors from the remaining 59 (65 total - 6 defective).

So, the number of ways to select this specific scenario is:

Number of ways to choose 3 defective transistors out of 6: C(6, 3)

Number of ways to choose 2 non-defective transistors out of 59: C(59, 2)

Using the combination formula again:

[tex]C(6, 3) \times C(59, 2) = (\frac{6!}{ (3! \times 3!)}) \times (\frac{59!} { (2! \times 57!)}) = 34220[/tex]

(c) Samples with no defective transistors simply means selecting all 5 transistors from the 59 non-defective ones.

Therefore:

[tex]C(59, 5) = \frac{59!}{ (5! \times 54!)} = 5006386[/tex]

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]

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: 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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When you have to repeatedly take the same test, with constant probability of succeeding/failing, you have to use Bernoulli's distribution. It states that, if you take [tex]n[/tex] tests with "succeeding" probability [tex]p[/tex], and you want to "succeed" k of those n times, the probability is

[tex]\displaystyle P(n,k,p) = \binom{n}{k}p^k(1-p)^{n-k}[/tex]

In your case, you have n=18 (the number of tests), and p=0.3 (the probability of succeeding). We want to succeed between 8 and 12 times, which means choosing k=8,9,10,11, or 12. For example, the probability of succeeding 8 times is

[tex]\displaystyle P(18,8,0.3) = \binom{18}{8}(0.3)^8(0.7)^{10}[/tex]

you can plug the different values of k to get the probabilities of succeeding 9, 10, 11 and 12 times, and your final answer will be

[tex]P = P(18,8,0.3) + P(18,9,0.3) + P(18,10,0.3) + P(18,11,0.3) + P(18,12,0.3)[/tex]

Final answer:

In mathematics, calculate the probability of a student guessing between 8 and 12 questions right out of 18 when the probability of guessing correctly is 0.3.

Explanation:

To calculate the probability of getting between 8 and 12 questions right through knowledgeable guessing, we can use the binomial probability formula:

P(k successes in n trials) = nCk × [tex]p^k[/tex] × (1-[tex]p^{(n-k)[/tex])

where:

k = number of successes (correct answers) = 8 to 12

n = total number of questions = 18

p = probability of success (correct answer by guessing) = 0.3

(1-p) = probability of failure (incorrect answer) = 0.7

We need to calculate the probability for each possible number of correct answers between 8 and 12 and then sum them up.

Here's the calculation:

Probability of getting 8 correct:

P(8 successes in 18 trials) = 18C8 ×0.3⁸ × 0.7¹⁰ ≈ 0.076

Probability of getting 9 correct:

P(9 successes in 18 trials) = 18C9 × 0.3⁹ ×0.7⁹ ≈ 0.124

Probability of getting 10 correct:

P(10 successes in 18 trials) = 18C10 ×0.3¹⁰ ×0.7⁸ ≈ 0.122

Probability of getting 11 correct:

P(11 successes in 18 trials) = 18C11 × 0.3¹¹ × 0.7⁷ ≈ 0.072

Probability of getting 12 correct:

P(12 successes in 18 trials) = 18C12 × 0.3¹² × 0.7⁶ ≈ 0.021

Total probability:

P(8 to 12 correct) = 0.076 + 0.124 + 0.122 + 0.072 + 0.021 ≈ 0.415

Therefore, the probability that the student gets between 8 and 12 questions right by knowledgeable guessing is approximately 41.5%.

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]

[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:

(d)

Step-by-step explanation:

Multiple your numbers out first. This always makes it less hectic to look at. Multiply both sides by x to move the x to the right side. Then divide both sides by ((0.4365)(87.9)).  You should get 10.

[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 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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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:

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]

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.