Let X ~ Exponential(). Show that

a. EX" = "EXn-1, for n = 1,2,3,...;
b. EX" = n!, for n = 1,2,3,....

The mass of a thin plate with varying density δ(x,y) occupying a region D in the plane xy is given by the double integral over D of δ(x,y) dA, where dA is an element of area in D.The mass of the plate is 21 units.

In this case, the plate covers the triangular region bounded by the x-axis and the lines x=1 and y=2x in the first quadrant, and the density of the plate is δ(x,y) = 6x + 6y + 6. Therefore, the mass of the plate is given by the double integral over the triangular region of (6x + 6y + 6) dA.

To evaluate this integral, we can use iterated integrals. First, we integrate with respect to y, keeping x constant:

∫[0,1] ∫[0,2x] (6x + 6y + 6) dy dx

This simplifies to:

∫[0,1] (18x + 12) dx

Integrating with respect to x, we get:

[tex](9x^2 + 12x) |_0^1 = 21[/tex]

Therefore, the mass of the plate is 21 units.

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The product rule must be used to find the number of ways to form this committee.

To find the number of ways to form a committee consisting of one representative from each of the 50 states in the United States, where the representative from a state is either the governor or one of the two senators from that state, we  must use the product rule. This is because for each state, there are three choices (governor or one of the two senators), and these choices are made independently for all 50 states.

So, you simply multiply the number of choices for each state together:

3 choices per state × 50 states = 3⁵⁰ possible ways to form the committee.

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The length of the rectangle is 46 meters, and the width is also 46 meters.

Let's start with the formula for the perimeter of a rectangle:

Perimeter = 2 x (length + width)

We know that the perimeter is 184 meters, so we can plug that in:

184 = 2 x (length + width)

Simplifying the equation, we get:

92 = length + width

To maximize the area, we need to find the dimensions that satisfy this equation and also maximize the area formula:

Area = length x width

We can use substitution to express the area in terms of one variable:

Area = length x (92 - length)

Now we have a quadratic equation that we can optimize using the vertex formula:

The x-value of the vertex of the parabola

[tex]y = ax^2 + bx + c[/tex] is given by -b/2a. In this case, a = -1 and b = 92.

The x-value of the vertex is -92/(-2) = 46.

The coefficient of the [tex]x^2[/tex] term is negative, the parabola is concave down and the vertex represents the maximum value of y.

The maximum area occurs when the length is 46 meters, and the width is:

width = 92 - length

width = 92 - 46

width = 46 meters

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Rounding to four decimal places, the probability is approximately 0.1293.

Given that the weights of steers in a herd are normally distributed with a mean (µ) of 1400 lbs and a variance (σ²) of 40,000 lbs², we first need to find the standard deviation (σ). We can do this using the formula:

σ = sqrt(σ²)

In this case, σ = sqrt(40,000) = 200 lbs.

Now, we need to find the z-scores for the weights 1580 lbs and 1720 lbs. The z-score formula is:

z = (X - µ) / σ

For 1580 lbs:

z1 = (1580 - 1400) / 200 = 0.9

For 1720 lbs:

z2 = (1720 - 1400) / 200 = 1.6

Next, we need to find the probability between these two z-scores. We can use a standard normal distribution table or calculator to find the probabilities corresponding to the z-scores:

P(z1) = P(Z ≤ 0.9) ≈ 0.8159
P(z2) = P(Z ≤ 1.6) ≈ 0.9452

Now, we'll subtract the probabilities to find the probability that the weight of a randomly selected steer is between 1580 and 1720 lbs:

P(0.9 ≤ Z ≤ 1.6) = P(Z ≤ 1.6) - P(Z ≤ 0.9) = 0.9452 - 0.8159 = 0.1293

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The regular hexagon has a area of 81 x √(3) square millimeters.

Draw a line from the center of the circle to one vertex of the hexagon, and draw another line from the center to the midpoint of a side of the hexagon.

This forms a right triangle, with the hypotenuse of length 12 millimeters, and one leg of length r, where r is the length of a side of the hexagon.

Using the Pythagorean theorem, we have:

r² + (r/2)² = 12²

Simplifying this equation, we get:

4r = 3 x 12

r = 12 x √(3)/2 = 6 x √(3)

Now that we have the length of the sides, we can use the formula for the area of a regular hexagon:

Area = (3 x √(3) / 2) x r²

Substituting the value of r, we get:

Area = (3 x √(3) / 2) x (6 x √(3))²

Area = (3 x √(3) / 2) x 108

Area = 81 x √(3) square millimeters

Therefore, the area of the regular hexagon is 81 x √(3) square millimeters.

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The p-value of the test statistic is 0.0512

We can conduct a hypothesis test for the proportion using a z-test.

The null hypothesis is that the proportion of readers who own a particular make of car is equal to 70%:

H0: p = 0.7

The alternative hypothesis is that the proportion is different from 70%:

Ha: p ≠ 0.7

The test statistic is calculated as:

z = (p' - p) / sqrt(p*(1-p)/n)

where p' is the sample proportion, p is the hypothesized proportion under the null hypothesis, and n is the sample size.

Plugging in the values from the problem, we get:

z = (0.64 - 0.7) / sqrt(0.7*(1-0.7)/200) = -1.96

Using a standard normal distribution table or calculator, we can find that the probability of getting a z-score of -1.96 or lower (or 1.96 or higher) is 0.0256. Since this is a two-tailed test, we double the probability to get the p-value:

p-value = 2*0.0256 = 0.0512

Therefore, the p-value of the test statistic is 0.0512, rounded to four decimal places.

Since the p-value is greater than the commonly used significance level of 0.05, we do not reject the null hypothesis and conclude that there is not enough evidence to support the claim that the proportion of readers who own a particular make of car is different from 70%.

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Answer:

5

Step-by-step explanation:

give me brainliest porfavor

The sum of p(z < z1) and p(z > z2) is  2 - 2p(z1 < z < z2)

We know that the probability of an event and its complement sum up to 1. That is,

p(A) + p(A') = 1

where A' is the complement of A. We can use this property to find the sum of p(z < z1) and p(z > z2) using the given information about p(z1 < z < z2).

Let A be the event {z < z1}, and B be the event {z > z2}. Then, A' is the event {z1 ≤ z ≤ z2}, and we have:

p(A') = p(z1 ≤ z ≤ z2) = p(z1 < z < z2)

Using the property that p(A) + p(A') = 1, we can solve for p(A):

p(A) = 1 - p(A') = 1 - p(z1 < z < z2)

Similarly, we can find p(B'):

p(B') = p(z1 < z < z2)

Using the complement property again, we have:

p(B) = 1 - p(B') = 1 - p(z1 < z < z2)

Therefore, the sum of p(z < z1) and p(z > z2) is:

p(z < z1) + p(z > z2) = p(A) + p(B) = 1 - p(z1 < z < z2) + 1 - p(z1 < z < z2) = 2 - 2p(z1 < z < z2)

Note that this result assumes that the distribution of z is symmetric about its mean.

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Answer:

for area, multiply the two sides and for perimeter, add the two sides and multiply the sum by 2.

Step-by-step explanation:

4, 6, and 8 are the three successive even numbers.

Call the smallest even integer "x" for simplicity. The following two even integers in succession would then be "x+2" and "x+4".

We may construct the following as an equation as the issue statement states that the product of the two smaller integers (x and x+2) is 16 greater than the largest integer (x+4):

x*(x+2) = (x+4) + 16

Extending the equation's left side results in:

x² + 2*x = x + 20

20 and x are subtracted from both sides to yield:

x² + x - 20 = 0

The factors of this quadratic equation are:

(x+5)(x-4) = 0

Consequently, x can have a value between -5 and 4. However, since we are aware that x must be an even number, the only viable answer is x = 4.

As a result, the three successive even integers are 4, 6, and 8; we can verify that this satisfies the initial requirement by checking that this:

4*6 = 24, which is 16 more than 8.

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The side length x measures approximately 8.60.

The figure in the image is a right triangle.

Angle θ = 35 degrees

Opposite to angle θ = x

Hypotensue = 15cm

The determine the side length x, we use the trigonometric ratio.

Note that:
sine = opposite / hypotenuse

Plug in the given values:

sin(35°) = x / 15

Cross multiply

x = sin(35°) × 15

x = 8.60

Therefore, the value of x is 8.60.

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Questions B, C, D, and E are all statistical questions.

Question B asks how many states Juanita has visited, which could be answered by counting the number of states she has visited.

Question C asks how many students are in Miss Lee's class today, which could be answered by counting the students in the class.

Question D asks how many students eat lunch in the cafeteria each day, which could be answered by counting the number of students who eat lunch in the cafeteria on a given day.

Question E asks how many pets each student at your school has at home, which could be answered by collecting data from each student about the number of pets they have at home.

Therefore, questions B, C, D, and E are all statistical questions.

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The number of sharks found between 76 m depth and 100 m depth is 2.04 shark's population.

The number of sharks found at different depths is calculated as follows;

From the histogram, the population of the fish at depth between 76 m and 100 m can be read off by tracing the depth values towards the frequency axis;

at depth 76 m, the frequency = 1.5

at depth 100 m, the frequency = 0.54

Total number of fish between the depths = 1.5 + 0.54 = 2.04

Thus, this value represents the density of the sharks between 76 m and 100 m.

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The equivalent gates for Figure P7.37 with bubbles on the input lines are (a) NOR gate and (b) AND gate.

(a) A bubble on the input of a gate represents inversion. In the case of (a) A B C = A + B, the bubble is on the output of the OR gate, indicating that the output is inverted. Thus, the equivalent gate is a NOR gate, which is an OR gate with an inverted output. The equation for the NOR gate is A B C = (A + B)'.

(b) Similarly, in (b) D E F = D E, the bubble is on the input of the AND gate, indicating that the input is inverted. Thus, the equivalent gate is an AND gate with an inverted input. The equation for the AND gate with an inverted input is D E F = D' E'.

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a. For a 95% confidence interval based on n = 15 observations, the degrees of freedom is 14. The t-value is 2.145. b. For a 90% confidence interval from an SRS of 24 observations, the degrees of freedom is 23. Using Table D, the t-value is 1.713. c. For a 95% confidence interval from a sample of size 24, the degrees of freedom is 23. The t-value is 2.069.

To find the t-value for each situation, we need to know the degrees of freedom, which is equal to n-1. Using this information, we can look up the t-value on Table D or use software to find the appropriate value.

a. For a 95% confidence interval based on n = 15 observations, the degrees of freedom is 14. Using Table D, the t-value is 2.145.
b. For a 90% confidence interval from an SRS of 24 observations, the degrees of freedom is 23. Using Table D, the t-value is 1.713.
c. For a 95% confidence interval from a sample of size 24, the degrees of freedom is 23. Using Table D, the t-value is 2.069.

These cases illustrate that as the sample size increases, the t-value decreases, which in turn reduces the size of the margin of error. Additionally, as the confidence level increases, the t-value increases, which increases the size of the margin of error. It is important to note that the size of the margin of error is also affected by the variability of the data, represented by the standard deviation.

To find the t-values for calculating the margin of error for a confidence interval for the mean of the population in the given situations, you can use software or a t-table (Table D) with the appropriate degrees of freedom and confidence level.

a. For a 95% confidence interval based on n = 15 observations, the degrees of freedom are 15-1 = 14. From Table D, the t-value (t*) is approximately 2.145.

b. For a 90% confidence interval from an SRS of 24 observations, the degrees of freedom are 24-1 = 23. From Table D, the t-value (t*) is approximately 1.714.

c. For a 95% confidence interval from a sample of size 24, the degrees of freedom are 24-1 = 23. From Table D, the t-value (t*) is approximately 2.069.

d. These cases illustrate that the size of the margin of error depends on the confidence level and the sample size. As the confidence level increases, the margin of error increases, and as the sample size increases, the margin of error decreases. This is because a higher confidence level requires a larger margin to ensure the true population mean falls within the interval, while a larger sample size provides more accurate estimates, reducing the margin of error.

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The coordinates of p(x)=55−12x−72x² relative to the basis B={4x²−3,3x−12+16x²,40−9x−52x²} in P₂ are [p(x)]_B = (12.48, -1.44, 0.475).

To find the coordinates of p(x) relative to the basis B, we first express p(x) as a linear combination of the basis elements in B. We then solve the resulting system of linear equations to find the values of the constants c1, c2, and c3.

Substituting these values into the expression for p(x) as a linear combination of the basis elements, we obtain the coordinates of p(x) relative to the basis B.

In this case, we found that c1=12-16c2+3c3, c2=-1.44, and c3=0.475, and thus [p(x)]_B=(12.48, -1.44, 0.475). This means that p(x) can be written as 12.48(4x²−3) -1.44(3x−12+16x²) + 0.475(40−9x−52x²) in terms of the basis B.

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Complete question:

The set B={4x  −3,3x−12+16x 2 ,40−9x−52x 2 } is a basis for P 2. Find the coordinates of p(x)=55−12x−72x 2relative to this basis: [p(x)] B=[:

Step-by-step explanation:

(PEMDAS).

50/(20-y)-1=4

50/(20-y)=5

Next, we can cross-multiply to get rid of the fraction:

50 = 5(20-y)

Simplifying further, we get:

50 = 100 - 5y

-50 = -5y

10 = y

Therefore, the solution to the equation is y=10.

Answer: y=10

The Equations have No solution.

We have,

y=5x+3

y=5x-2

Solving the Equation we get

5x + 3 = 5x - 2

5x - 5x = -2-3

0 = -5

Also, 5/5 = -1 / (-1) ≠ 3-/2

Thus, the equation have no solution as the equation represent parallel line.

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Here's a proof for each of the statements you provided.

a) If g∘f = I_A, then f is an injection.
Proof: Assume x1 and x2 are elements of A such that f(x1) = f(x2). We want to show that x1 = x2. Since g∘f = I_A, we have g(f(x1)) = g(f(x2)). Applying I_A, we get x1 = g(f(x1)) = g(f(x2)) = x2. Thus, f is injective.

b) If f∘g = I_B, then f is a surjection.
Proof: Let y be an element of B. We want to show that there exists an element x in A such that f(x) = y. Since f∘g = I_B, we have f(g(y)) = I_B(y) = y. Thus, there exists an element x = g(y) in A such that f(x) = y. Therefore, f is surjective.

c) If g∘f = I_A and f∘g = I_B, then f and g are bijections and g = f^(-1).
Proof: From parts (a) and (b), we know that f is both injective and surjective, which means f is a bijection. Similarly, g is also a bijection. Now, we need to show that g = f^(-1). By definition, f^(-1)∘f = I_A and f∘f^(-1) = I_B. Since g∘f = I_A and f∘g = I_B, it follows that g = f^(-1).

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The calculated value of the measure of the angle V is 81 degrees

From the question, we have the following parameters that can be used in our computation:

Because the triangle ERT is congruent to triangle CVB, then we have

E = C

So, we have

7x + 4 = 32

Evaluate the like terms

7x = 28

Divide by 7

x = 4

Also, we have

V = R

This means that

V = 180 - C - B

Substitute the known values in the above equation, so, we have the following representation

V = 180 - 7x - 4 - 15x - 7

So, we have

V = 180 - 7(4) - 4 - 15(4) - 7

Evaluate

V = 81

Hence, the measure of the angle is 81 degrees

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An outlier in a small sample size can have a significant effect on a confidence interval. It can cause the interval to widen, leading to increased uncertainty and decreased precision in estimating the population parameter.

Confidence intervals are statistical ranges used to estimate population parameters based on sample data. In small sample sizes, each data point has a greater impact on the overall result.

An outlier, which is a data point significantly different from the rest of the sample, can distort the calculations used to construct the confidence interval. Since the interval aims to capture the true population parameter with a specified level of confidence, the presence of an outlier can lead to increased variability in the data.

As a result, the confidence interval may need to be widened to account for the potential influence of the outlier, reducing the precision and increasing the uncertainty in estimating the parameter. Therefore, outliers can have a notable effect on confidence intervals based on small sample sizes.

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The area of the lake on the map scale of  1:200000 is found to be 0.496 km²

To find out the size of the lake, we have to find the area of the map and then we will use the scaling.

We know that the scale of the map is 1:200000. This means that 1 centimeter on the map represents 200,000 centimeters on the ground. Now, converting the values. So, the area of the lake on the ground is,

(12.4cm²/10,000,000)(200,000cm/1cm)² = 0.496 km²

Therefore, the area of the lake is 0.496 square kilometers (rounded to three decimal places).

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The correct answer is A. 3 mph. If Isaac's walking rate remains constant, so Isaac's walking rate is 3 mph

To find the Isaac's walking rate in miles per hour, we exactly need to divide the distance which he walks by the time it takes for him to walk that similar distance. We are given that the Isaac walks 6/10 of a mile in 1/5 of an hour, so:

Walking rate = distance ÷ time

Walking rate = (6/10) ÷ (1/5)

Walking rate = (6/10) x (5/1)

Walking rate = 3 miles per hour

Therefore, Isaac's walking rate is 3 mph.

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The difference between the theoretical and experimental probability of rolling a sum of 4 with a pair of dice based on Tim's experiment is -11/300.

To find the difference between the theoretical and experimental probability of rolling a sum of 4 with a pair of dice based on Tim's experiment, we first need to determine both probabilities.

The theoretical probability can be calculated as follows:
1. There are a total of 6x6=36 possible outcomes when rolling two dice.
2. The combinations that result in a sum of 4 are (1, 3), (2, 2), and (3, 1).
3. There are 3 favorable outcomes for a sum of 4, so the theoretical probability is 3/36, which simplifies to 1/12.

The experimental probability is based on Tim's experiment, where he rolled the dice 25 times:
1. He recorded a sum of 4 on three of those rolls.
2. The experimental probability is the number of successful outcomes (rolling a 4) divided by the total number of trials (25 rolls). So, the experimental probability is 3/25.

Finally, find the difference between the theoretical and experimental probability:
1. The theoretical probability is 1/12, and the experimental probability is 3/25.
2. To compare them, find a common denominator (which is 300) and convert both probabilities: (25/300) - (36/300).
3. Subtract the probabilities: 25/300 - 36/300 = -11/300.

The difference between the theoretical and experimental probability of rolling a sum of 4 with a pair of dice based on Tim's experiment is -11/300.

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Answer:

x = 14

Step-by-step explanation:

using the cosine ratio in the right triangle and the exact value

cos30° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

cos30° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{7\sqrt{3} }{x}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

x × [tex]\sqrt{3}[/tex] = 14[tex]\sqrt{3}[/tex] ( divide both sides by [tex]\sqrt{3}[/tex] )

x = 14

To determine the probability of a student getting exactly 10 questions correct, we need to know the total number of ways in which the student can answer the questions and the number of ways in which the student can get exactly 10 questions correct.

Assuming that each question has only two possible answers (e.g. true/false or multiple choice with two options), and the student guesses randomly, the probability of getting a single question correct is 1/2, and the probability of getting a single question incorrect is also 1/2.

Let X be the number of questions the student answers correctly, and n be the total number of questions.

In this case, n = 20 (the total number of questions), and p = 1/2 (the probability of getting a single question correct).

where (n choose k) is the binomial coefficient, which represents the number of ways to choose k items from a set of n items, and is given by:

(n choose k) = n! / (k! * (n - k)!)

P(X = 10) = (20 choose 10) * (1/2)^10 * (1/2)^(20-10)

= 184,756 * 0.0009765625 * 0.0009765625

= 0.1801

Therefore, the probability of the student getting exactly 10 questions correct is 0.1801, or approximately 18.01%.

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1. There is no solution.

2. The two positive numbers with product 200 such that the sum of one number and twice the second number is as small as possible are 40 and 5.

3. The length and width of the rectangle should be 25 cm and 50 cm, respectively, so that the area is a maximum.

1. Let the two numbers be x and y, where x > y. We have the equation:

x - y = 250

This can be rearranged to give:

x = y + 250

The product of the two numbers is given by:

P = xy = y(y + 250) = y^2 + 250y

To find the minimum value of P, we take the derivative with respect to y and set it equal to zero:

dP/dy = 2y + 250 = 0

Solving for y, we get:

y = -125

Since we require positive numbers, this is not a valid solution. Therefore, we take the second derivative:

d^2P/dy^2 = 2 > 0

This confirms that we have a minimum. To find the corresponding value of x, we use the equation x = y + 250:

x = -125 + 250 = 125

Therefore, the two numbers are 125 and -125, but since we require positive numbers, there is no solution.

2. Let the two numbers be x and y, where x > y. We are given that:

xy = 200

We want to minimize the expression:

x + 2y

We can solve for one variable in terms of the other:

x = 200/y

Substituting into the expression to be minimized, we get:

x + 2y = 200/y + 2y = 200/y + 4y/2 = 200/y + 2y^2/y

Simplifying, we get:

x + 2y = (200 + 2y^2)/y

To minimize this expression, we take the derivative with respect to y and set it equal to zero:

d/dy (200 + 2y^2)/y = -200/y^2 + 4y/y^2 = 4y/y^2 - 200/y^2 = 0

Solving for y, we get:

y = 5

Substituting back into the equation xy = 200, we get:

x = 40

Therefore, the two positive numbers with product 200 such that the sum of one number and twice the second number is as small as possible are 40 and 5.

3. Let the length and width of the rectangle be x and y, respectively. We are given that the perimeter is 100, so:

2x + 2y = 100

Solving for y, we get:

y = 50 - x

The area of the rectangle is given by:

A = xy = x(50 - x)

To maximize this expression, we take the derivative with respect to x and set it equal to zero:

dA/dx = 50 - 2x = 0

Solving for x, we get:

x = 25

Substituting back into the equation y = 50 - x, we get:

y = 25

Therefore, the length and width of the rectangle should be 25 cm and 50 cm, respectively, so that the area is a maximum.

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It is proved that AK = LA when two distinct nonparallel lines k  and l are tangent to C(0,r).

Since k and l are nonparallel, they must intersect at some point P. Let Q be the center of the circle C(0,r). Then, by the tangent-chord angle theorem, we have:

∠AKP = 90°   (since k is tangent to the circle at K)

∠ALP = 90°   (since l is tangent to the circle at L)

Also, by the angle between intersecting lines, we have:

∠KPL = ∠APK + ∠APL

Since k and l are tangent to the circle at K and L, respectively, we have:

∠KQL = ∠PQK = 90°   (tangent-chord angle theorem)

∠LQK = ∠PQL = 90°   (tangent-chord angle theorem)

Therefore, quadrilateral KQLP is a rectangle, and we have:

∠KPL = 180° - ∠KQL - ∠LQK = 180° - 90° - 90° = 0°

This means that points K, P, and L are collinear, so we can write:

KP + PL = KL

Since k and l are tangent to the circle, we have:

KP = PL = r

Therefore, we have:

AK + AL = KL = 2r

It remains to show that AK = AL. Suppose, for the sake of contradiction, that AK ≠ AL. Without loss of generality, assume that AK < AL. Then we have:

AK + AL = 2AK + (AL - AK) < 2AL = KL

This contradicts the fact that AK + AL = KL, so our assumption must be false. Therefore, we conclude that AK = AL.

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If you toss a fair coin repeatedly, the probability of getting a head in each individual toss is 1/2. Therefore, the probability of not getting a head in the first toss is 1/2, in the first two tosses is (1/2)^2, and so on. We can define the events An = {"no head in the first n tosses"}. The probability of An is (1/2)^n for any n.

Using the complement rule, we can say that the probability of getting a head in the first n tosses is 1 - (1/2)^n.
Now, we can consider the infinite sequence of events {A1, A2, A3, ...}. By the union bound, the probability of not getting a head in any of the tosses is the probability of An for all n.

This can be expressed as the infinite product of (1/2)^n, which is 0. Therefore, the probability of getting a head eventually is 1.

In simpler terms, even though the probability of getting a head on any individual toss is 1/2, if you keep tossing the coin, the probability of never getting a head decreases exponentially. So, with probability one, you will eventually get a head.

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