Given P(A and B) 0.20, P(A) 0.49, and P(B) = 0.41 are events A and B independent or dependent? 1) Dependent 2) Independent

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:

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:

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:

  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:

  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

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]

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.

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:

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:

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: 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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[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]

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 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:

(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.

Answer:

The  value of his investment in 3 years = $10007.53

Step-by-step explanation:

Compound interest

A = P[1 +R/n]^nt

Where A - amount

P - principle amount

R = rate of interest

t -  number of years

n - number of times compounded yearly

To find value if investment

Here P = $9500,  R = 1.75% = 0.0175 n = 1 and t = 3

A = P[1 +R/n]^nt

 = 9500[1 + 0.0175/1]^(1 * 3)

 = 10007.53

Answer:

$10,012.07

Step-by-step explanation:

Identify the values of each variable in the formula. Remember to express the percent as a decimal.

APrt=?=$9,500=0.0175=3 years

For compounding continuously, use the formula A=Pert.

Substitute the values into the formula and compute the amount to find

AA=9,500e0.0175⋅3=$10,012.07.

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:

The slope is: 7

The y-intercept is: 1

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

[tex]y=mx+b[/tex]

Where "m" is the slope of the line and "b" is the y-intercept.

To find the slope and the y-intercept of the given line, we can write it in Slope-Intercept form. We can do this by solving for "y".

Then, this is:

[tex]-7x + y = 1\\\\y=7x+1[/tex]

Therefore, you can identify that the slope of this line is:

[tex]m=7[/tex]

And the y-intercept is:

[tex]b=1[/tex]

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:

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]

[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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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:

  c)  M/40

Step-by-step explanation:

The share of each company was originally M/10. Now it is M/8. The increase is ...

  M/8 -M/10 = M(1/8 -1/10) = M(10-8)/(10·8) = M/40

The increase in the share of each company after two companies dropped out is M/40.

The total amount of money to be paid by the companies is M rupees. Initially, these M rupees were to be divided among 10 companies, but when two dropped out, they had to be divided among 8 companies instead. So, the share of each company initially was M/10 and after the drop out, the share of each company became M/8.

The increase in the amount that each company had to give is given by subtraction of the initial share from the final share, that is, (M/8) - (M/10). To solve this equation, we first must find a common denominator for the fractions, which is 40 in this case. Therefore, the equation becomes (5M/40) - (4M/40) = M/40.

So, the increase in share of each company was M/40. Therefore, the correct answer is (c) M/40.

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So we have a set of integers,

[tex]X=\{15,20,25,25,27,28,30,34\}[/tex]

And we need to compute some stats.

First the range,

The range of data set is difference between the maximum and the minimum of the data set.

So the minimum is 15 and maximum is 34 therefore the range is 34 - 15 = 19.

Second the interquartile range,

The interquartile range of data set is the difference of the first and third quartiles.

First quartile is the value separating the lower quarter and higher three - quarters of the set. The first quartile is computed by taking median of the lower half of the sorted set. This is 15, 20, 25, 25 and it's median is 22.5

Third quartile is therefore the median of 27, 28, 30, 34 which is 29.

Next up, variance.

The variance of set measures how much data is spread out. For data set [tex]x_1,\dots, x_n[/tex] with an average [tex]a[/tex],

[tex]\mathsf{var}(X)=\Sigma_{i=1}^{n}\dfrac{(x_i-a)^2}{n-1}[/tex]

So first compute the average value which is actually a mean,

[tex]a=\dfrac{1}{2}\Sigma_{i=1}^{n}a_i[/tex]

The sum [tex]\Sigma_{i=1}^{n}a_i[/tex] of numbers in set is 204.

Divide this by number of elements in the set (8).

[tex]a=\dfrac{204}{8}=25.5[/tex]

Then compute the variance,

[tex]\mathsf{var}(X)=\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}\approx\boxed{34.57}[/tex]

And finally standard deviation,

Since have computed variance the standard deviation is nothing too hard. It's defined as a square root of variance,

[tex]\mathsf{sde}(X)=\sqrt{\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}}=\sqrt{34.57}\approx\boxed{5.88}[/tex]

Hope this helps.

r3t40

The values of the statistical measures for the distribution are :

Given the data :

27, 25, 20, 15, 30, 34, 28, 25.

Rearranging the values :

15, 20, 25, 25, 27, 28, 30, 34

A.)

The range = Maximum - Minimum

Maximum = 34 ; Minimum = 15

Range = 34 - 15 = 19

B.)

The interquartile range = Upper quartile - Lower quartile

Upper quartile (Q3) = 3/4(n+1)th term

n = number of values = 8

Q3 = 3/4(9) = 6.75 th term = (28+30)/2 = 29

Lower quartile (Q1) = 1/4(n+1)th term

Q1 = 1/4(9) = 2.25th term = (20+25)/2 = 22.5

Interquartile range = 29 - 22.5 = 6.5

C.)

Variance :

Σ[(X - mean)²] / (n-1)

ΣX/ n

Mean = 204 / 8 = 25.5

[(15-25.5)² + (20-25.5)² + (25-25.5)² + (25+25.5)² + (27-25.5)² + (28-25.5)² + (30-25.5)² + (34 - 25.5)²] ÷ (8 - 1)

Variance = 242 / 7

Variance = 34.57

Standard deviation = √variance

Standard deviation = √34.57

Standard deviation = 5.879

Standard deviation = 5.88 (2 decimal place).

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Answer:

x=10 and W=12

Step-by-step explanation:

Let's solve the equations. First we need to understand that the problem can be solved because we have two variables (x, W) and two equations.

Now, we have the following equations:

3x+3W-66 making the equation equal to 0:

3x+3W-66=0 which can be express as:

3x=-3W+66

x=(-3W+66)/3

x=-W+22 (equation 1)

The next equation is:

12x+15W-300 making the equation equal to 0 and then divided by 3:

(12x+15W-300)/3=0 which is:

4x+5W-100=0 (equation 2), using equation 1 we can write:

4(-W+22)+5W-100=0

-4W+88+5W-100=0

W-12=0

W=12

Using W=12 in equation 2 we have:

4x+5W-100=0

4x+5*(12)-100=0

4x+(60)-100=0

4x-40=0

4x=40

x=40/4

x=10

In conclusion the solution for the equations are: x=10 and W=12.

[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: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:

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:

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 ..