row equivalent matrix method
4x-3y=11
3x+7y=-1​

the dimensions of the rectangle are L = 15 inches and W = 15 inches, and the minimum perimeter is:    P = 2L + 2W = 60 inches.

Let the length and width of the rectangle be L and W, respectively, so that the area of the rectangle is A = LW = 225. We want to find the dimensions of the rectangle with minimum perimeter P = 2L + 2W, and then find the minimum perimeter.

Using the given area, we can solve for one of the variables in terms of the other:

L = 225/W

Substituting this expression for L into the expression for the perimeter, we get:

P = 2(225/W) + 2W

Taking the derivative of P with respect to W and setting it equal to zero to find the minimum, we get:

[tex]dP/dW = -450/W^2 + 2 = 0[/tex]

Solving for W, we get:

W^2 = 225

Since W must be positive (it is a length), we take the positive square root:

W = 15

Substituting this value of W back into the expression for L, we get:

L = 225/W = 15

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Since there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

To show that S is uncountable, we need to show that there is no one-to-one correspondence between the set of infinite sequences of 0s and 1s and the set of natural numbers (which are used to count elements in a set).
We can use the diagonal argument, which is similar to the one used to show that the real numbers are uncountable.
Assume that S is countable, meaning that there is a way to list all the infinite sequences of 0s and 1s as a sequence: s1, s2, s3, ...
We can construct a new sequence t by choosing the opposite of each digit on the diagonal of the previous sequences: if the diagonal digit of s1 is 0, then the first digit of t is 1, and vice versa. Similarly, if the diagonal digit of s2 is 1, then the second digit of t is 0, and vice versa. And so on for all the diagonal digits.
The sequence t is guaranteed to be different from all the sequences in the original list, because it differs from each sequence at least at one position (the diagonal position). Therefore, t cannot be in the original list, which means that the original list did not include all the infinite sequences of 0s and 1s.
Since we have shown that there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

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Answer: x= 2, -2.5

Step-by-step explanation: trust lol

The equation to show the number of cakes that can be baked with 50 pounds of flour, is 181. 5 = c × 112.

If we represent the quantity of cakes Betty can make as "c", and it is known that each cake requires 112 cups of flour, with a total of 181.5 cups available, then the equation may be expressed as:

Total flour = Number of cakes × Flour per cake

181. 5 = c × 112

c = 181. 5 / 112

c = 1.62 cakes

In conclusion, 1. 62 cakes can be baked.

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The vertex that is not part of the image is:

A:(-4,6)

The translations to an image are represented as follows:

The vector notation of a translation is given as follows:

{x ± a, y ± a}

We have the next polygon is a translation of the first along the vector (3,-3).

The rule applied to each vertex of the image is:

(x, y) → (x + 3, y - 3).

Now, We have the vertices are:

(-7,3),(-4,6),(-1,3) and (-4,0).

Applying the translation rule, the vertices of the image are as follows:

(-4, 0), (-1, 3), (2, 0), (-1, -3).

The lone coordinate without a vertex of the image is (-4,6)

The correct option is (A)

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The segment that is opposite ∠ E is C. UJ.

A segments that could be formed through connecting the endpoints of the adjacent segments are called the opposite segments. This can also create an intersection in certain types of lines or segment configurations, depending on their individual set ups.

When looking at angle ∠ E, we can see that the segment opposite it is Segment UJ. This is because it was formed by connecting the endpoints of the adjacent segments of UE and JE.

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Every neighborhood of A contains a point of B and every neighborhood of B contains a point of A, which implies that A=B.

To show that there exists a sequence of rational numbers converging to any real number x, we can use the fact that the rational numbers are dense in the real numbers. This means that between any two real numbers, there exists a rational number.

So, let x be any real number. We can construct a sequence of rational numbers {q_n} such that q_n is the rational number between x-1/n and x+1/n. In other words,

q_n = a/b, where a and b are integers such that x-1/n < a/b < x+1/n and b > n

Then, it can be shown that as n approaches infinity, q_n converges to x. Therefore, there exists a sequence of rational numbers converging to any real number x.

To prove that A=B, we need to show that every neighborhood of A contains a point of B and every neighborhood of B contains a point of A.

First, let's consider any neighborhood of A. Since {a_n} converges to A, we know that there exists some positive integer N such that for all n > N, |a_n - A| < ε/2, where ε is the radius of the neighborhood.

Now, since B is an accumulation point of {a_n : n ∈ J}, we know that there exists some integer j ∈ J such that |a_j - B| < ε/2.

Thus, we have:

|A - B| ≤ |A - a_j| + |a_j - B| < ε/2 + ε/2 = ε

This shows that B is also in the neighborhood of A.

Next, let's consider any neighborhood of B. Since B is an accumulation point of {a_n : n ∈ J}, we know that there exists some positive integer M such that there are infinitely many n ∈ J satisfying |a_n - B| < ε/2.

Now, let n_1, n_2, n_3, ... be a subsequence of {a_n} such that |a_ni - B| < ε/2 for all i ≥ 1.

Since {a_n} converges to A, we know that there exists some positive integer N such that for all n > N, |a_n - A| < ε/2.

Let N' be the maximum of N and n_1, so that for all n > N', we have:

|a_n - A| < ε/2 and |a_n - B| < ε/2

Then, we have:

|A - B| ≤ |A - a_n| + |a_n - B| < ε/2 + ε/2 = ε

This shows that A is also in the neighborhood of B.

Therefore, we have shown that every neighborhood of A contains a point of B and every neighborhood of B contains a point of A, which implies that A=B.

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The given procedure does result in a binomial distribution. This is because it meets the necessary conditions for a binomial distribution:

1. Fixed number of trials: There are 74 trials (rolling the die 74 times).
2. Two outcomes: The outcome of each roll is either odd (success) or even (failure).
3. Independent trials: The outcome of one roll does not affect the outcome of any other roll.
4. Constant probability: The probability of rolling an odd number remains the same for each roll (1/2, since there are 3 odd numbers out of 6 sides).'

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Answer: Whole thing=20.5

First half (17,18,19,20)=18.5

Second half (21,24,25,27) = 24.5

Step-by-step explanation:

eliminate numbers on both sides till you get to the middle if there is an even number add up the two numbers in the middle and divide them by 2

For example, In the whole thing, you are left with 20 and 21 so

20+21 =41/2= 20.5

For example, In the first part, you are left with 18 and 19 so

18+19 =37/2= 18.5

For example, In the first part, you are left with 24 and 25 so

24+25 =49/2= 24.5

The value of x in the right triangle with acute angles 8x and 4x + 6 is 7

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

The right triangle with acute angles 8x and 4x + 6

The sum of acute angles in a right triangle is 90

Using the above as a guide, we have the following:

8x + 4x + 6 = 90

So, we have

12x = 84

Divide by 12

x = 7

Hence, the value of x is 7

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The relation that has the  domain {-3, 4} and the range {0, 1} is B:

{( −3, 1), (4, 0)}

Remember that for any relation, the domain is the set of the inputs and the range is the set of the outputs, and the general notation for a point is (input, output).

Then if the domain is {-3, 4} and the range is {0, 1} the only of the given relations that can be described by these is:

B. {( −3, 1), (4, 0)}

So that is the correct option.

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Therefore, the apparent nth term of the expression is: aⁿ = 5n² - 3n - 2.

The given sequence is not an arithmetic or geometric sequence. However, we can notice that the sequence of differences between consecutive terms is an arithmetic sequence.

The sequence of differences is: 7, 19, 37, 61,...

To find the nth term of this sequence, we can use the formula for the nth term of an arithmetic sequence:

dn = a1 + (n-1) * d

where dn is the nth term of the sequence of differences, a1 is the first term of the sequence of differences, d is the common difference of the sequence of differences, and n is the index of the term we want to find.

So, we have:

dn = 7 + (n-1) * 12

Simplifying this expression, we get:

dn = 5n - 3

Now, we can use this formula to find the nth term of the original sequence. Let's call the nth term an:

an = an-1 + dn-1

where an-1 is the (n-1)th term of the original sequence and dn-1 is the (n-1)th term of the sequence of differences.

We know that a1 = 2 and d1 = 7, so we can use the above formula to find the next terms:

a2 = a1 + d1 = 2 + 7 = 9

a3 = a2 + d2 = 9 + 19 = 28

a4 = a3 + d3 = 28 + 37 = 65

a5 = a4 + d4 = 65 + 61 = 126

Therefore, the apparent nth term of the sequence is: aⁿ = 5n² - 3n - 2.

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The sales of Indiana's businesses during the spring of 2020 are modeled by the equation Sales = 500(1-0.10)^t, where t is the number of days and sales are in millions of dollars. If sales reach $23.5 million, it will be considered a statewide economic crisis.

To solve the problem, we need to use the given equation and substitute the value of sales ($23.5 million) into it. Then we can solve for the value of t, which represents the number of days until sales reach the economic crisis.

500(1-0.10)^t = 23.5

(1-0.10)^t = 0.047

Taking the natural logarithm of both sides,

ln[(1-0.10)^t] = ln(0.047)

t ln(0.90) = -3.057

t = -3.057 / ln(0.90)

Using a calculator, we can evaluate the right-hand side of the equation to get t ≈ 37.28 days.

Therefore, it will take approximately 37.28 days for the sales of Indiana's businesses to reach the economic crisis threshold of $23.5 million.

In summary, we used the given exponential equation to find the number of days until the sales of Indiana's businesses reach the economic crisis threshold of $23.5 million. By substituting the value of sales into the equation and solving for t, we found that it will take approximately 37.28 days for this critical point to be reached. This calculation highlights the impact of the COVID-19 pandemic on the state's economy and underscores the importance of economic stimulus measures during times of crisis.

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If you provide more values than the size of the array, you'll get a compilation error.

In C or C++ programming languages, an array can be declared, created, and initialized in a single statement. Here's how you can declare, create, and initialize an array named a of 10 elements of type int with the values of the elements (starting with the first) set to 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100, respectively:

int a[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};

This statement does the following:

Declares an array named a of 10 elements of type int.

Initializes the elements of the array with the specified values in the curly braces, starting from the first element.

Note that if you don't provide enough values in the curly braces, the remaining elements will be initialized to 0. If you provide more values than the size of the array, you'll get a compilation error.

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A. y = 4x-7, the inverse function is, x = (y + 7)/4

B. y = 7x, the inverse function is, x = y/7

C. y = 7x-4, the inverse function is, x = (y + 4)/7

D. y = x-4, the inverse function is, x = y + 4

E. y = x+4, the inverse function is, x = y - 4

F. y = x-4/7, the inverse function is, x = 7y + 4

The inverse of each of the functions is calculated as follows;

y = 4x - 7

x = 4y - 7

4y = x + 7

y = (x + 7)/4  

⇒ x = (y + 7)/4

Second function;

y = 7x

x = 7y

y = x/7

⇒ x = y/7

Third function;

y = 7x - 4

x = 7y - 4

7y = x + 4

y = (x + 4)/7

⇒ x = (y + 4)/7

Fourth function;

y = x - 4

x = y - 4

y = x + 4

⇒ x = y + 4

Fifth function;

y = x + 4

x = y + 4

y = x - 4

⇒ x = y - 4

Sixth function;

y = (x - 4)/7

x = (y - 4)/7

7x = y - 4

y = 7x + 4

⇒ x = 7y + 4

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Answer: #3 = 22/50 (reduced version is 11/25

Step-by-step explanation:

Answer:

Step-by-step explanation:

Predicted probabilities are different then experimental probabilities.  Experimental probabilities use the actual data.

P(red)= red/(total)= 22/(12+15+22)  = 22/50 = .44

P(hot cocoa = hot cocoa/total =5/(7+5+8) = 1/4 = .25

In the given triangle, the measure of angle A is approximately 55°

From the question, we are to determine the measure of angle A

To determine the measure of angle A, we will use SOH CAH TOA

sin (angle) = Opposite / Hypotenuse

cos (angle) = Adjacent / Hypotenuse

tan (angle) = Opposite / Adjacent

Thus,

We can write that

sin (A) = BC / AB

First, we will determine the length of BC

From the Pythagorean theorem,

BC² = AB² - AC²

BC² = 37² - 21²

BC² = 928

BC = √928

BC = 4√58

Thus,

sin (A) = (4√58) / 37

sin (A) = 0.8233

A = sin⁻¹ (0.8233)

A = 55.4165°

A ≈ 55°

Hence,

The measure of angle A is 55°

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We can be 96% confident that the true mean weight of all women lies between 129.21 and 367.11.

The best point estimate of the mean weight of all women can be calculated using the formula:

Point estimate = sample mean = (sum of sample weights) / sample size

Here, the sample size is not given, so we cannot calculate the sample mean directly. However, we are given two sample statistics: the sample starting point (2) and the sample statistic (s) which is the sample standard deviation.

We can use the formula for the t-distribution to estimate the population mean:

t = (x - μ) / (s / √n)

where x is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.

To find the point estimate, we can rearrange this formula to solve for x:

x = μ + t(s / √n)

Since we don't know the population mean μ, we will use the sample starting point 2 as an estimate. We also know the sample standard deviation s = 30.766 and we are given a 96% confidence interval, so we need to find the critical value of t for a two-tailed test with 96% confidence and degrees of freedom (df) = n - 1.

Using a t-distribution table or calculator, we find that the critical value for df = n - 1 = 1 is t = 12.71.

Plugging in the values, we get:

2 + 12.71 * (30.766 / √n) = x

Solving for x, we get:

x = 2 + 12.71 * (30.766 / √n)

We still need to find the sample size n in order to calculate the point estimate. We can use the sample statistic given, which is the sample standard deviation s = 30.766, to estimate the sample size using the formula:

s = √[(n-1)/n] * σ

where σ is the population standard deviation.

Plugging in the values, we get:

30.766 = √[(n-1)/n] * 30.766

Solving for n, we get:

n = 2.24

This suggests that the sample size is quite small, which may limit the accuracy of our point estimate.

Plugging in the value of n, we get:

x = 2 + 12.71 * (30.766 / √2.24)

x = 2 + 12.71 * 19.398

x = 248.16

Therefore, the best point estimate of the mean weight of all women is 248.16.

b. To find a 96% confidence interval for the mean weight of all women, we can use the formula:

CI = x ± t(α/2, df) * (s / √n)

where x is the point estimate, t(α/2, df) is the critical value for a two-tailed test with α = 0.04 and df = n - 1, s is the sample standard deviation, and n is the sample size.

Plugging in the values, we get:

CI = 248.16 ± 12.71 * (30.766 / √2.24)

CI = 248.16 ± 118.95

CI = (129.21, 367.11)

Therefore, we can be 96% confident that the true mean weight of all women lies between 129.21 and 367.11.

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Using  both and levels of significance and the critical value, we can conclude that the percentage of spam emails sent by the huge firm is different from the percentage in a recent month.

The population proportion of spam emails in a recent month is p = 0.718.

A random sample of emails from a large corporation has a sample proportion of spam emails, p'= 0.645.

We want to test the hypothesis that the population proportion of spam emails in the large corporation is different from p = 0.718.

We will use both 0.05 and 0.01 levels of significance and the critical value method.

To test this hypothesis using the critical value method, we can follow these steps:

The null hypothesis is that the population proportion of spam emails in the large corporation is equal to 0.718:

H0: p = 0.718

The alternative hypothesis is that the population proportion of spam emails in the large corporation is different from 0.718:

Ha: p ≠ 0.718

We will use both 0.05 and 0.01 levels of significance. Since we have a large sample (np > 10 and n(1-p) > 10), we can use the z-test for proportions. The test statistic is calculated as:

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

where n is the sample size.

Using a standard normal distribution table, the critical values for a two-tailed test at the 0.05 and 0.01 levels of significance are:

At the 0.05 level: ±1.96

At the 0.01 level: ±2.58

Step 4: Calculate the test statistic and p-value.

Using the formula for the test statistic and the given values, we get:

z = (0.645 - 0.718) / sqrt(0.718(1-0.718)/n)

Since we don't know the population standard deviation, we use the standard error estimated from the sample:

z = (0.645 - 0.718) / sqrt(0.718(1-0.718)/n) = -2.546 / sqrt(0.718(1-0.718)/n)

Using n = 1000 (a reasonable sample size for an email server), we get:

z = -2.546 / sqrt(0.718(1-0.718)/1000) = -9.386

The corresponding p-value for this test statistic is very small (less than 0.0001), indicating strong evidence against the null hypothesis.

At the 0.05 level of significance, the critical value is ±1.96, which does not include the calculated test statistic of -9.386. Therefore, we reject the null hypothesis and conclude that the population proportion of spam emails in the large corporation is different from 0.718.

At the 0.01 level of significance, the critical value is ±2.58, which also does not include the calculated test statistic of -9.386. Therefore, we reject the null hypothesis at this level of significance as well.

In conclusion, we have strong evidence to suggest that the proportion of spam emails in the large corporation is different from the proportion in a recent month (0.718).

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The statement "The normal force is equal to the perpendicular component of the object's weight, which decreases as the angle of inclination increases" is true.

As the angle of inclination increases, the object's weight can be divided into two components: one perpendicular to the inclined surface (the normal force) and one parallel to it. As the angle increases, the perpendicular component (normal force) decreases, while the parallel component increases.

So to directly answer your question, the normal force is never equal to the weight of the object on an inclined plane (unless you count the limiting case of level ground). It is equal to the weight of the object times the cosine of the angle the inclined plane makes with the horizontal.

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From percentage formula, in a vine gunshot sound effect download unblocked restaurants, the number of expected Italian restaurants in next 24 menus slipped under Eli's door is six.

We have a Vine gunshot sound effect download unblocked Restaurants often slip takeout menus. Total collected menus by Eli = 28

In this 28 menus, 7 for Italian food. We have to determine the how many menus in next 24 menus slipped under Eli's door expected from Italian restaurants. Percentage is calculated by dividing the value by the total value, and then multiplying the resultant value by 100.

Using percentage formula, 7 are Italian food out of all 28 that is total [tex]\frac{7 {28} × 100 = 25\%[/tex].

Let the required number of menus slipped under Eli's door should you expect to be from Italian restaurants be x. So, number of menus slipped under Eli's door should you expect to be from Italian restaurants out of 24, x = 25% of 24

=> [tex]\frac{25}{100} × 24 = x[/tex]

=> x = 6

Hence, required value is 6.

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Using the stars and bars formula, the number of solution sets for a+b+c+d+e = 18 is 7315, which is obtained by arranging 18 stars and 4 bars in a line, giving a total of 22 objects, and choosing 4 of them to be the bars.

This problem can be solved using the "stars and bars" combinatorial technique. We can think of 18 stars representing the total sum, and 4 bars dividing them into 5 bins.

There are a total of 22 objects (18 stars and 4 bars), and we need to choose the positions of the 4 bars out of the 22 objects, which can be done in (22 choose 4) ways.

Therefore, there are (22 choose 4) = 7315 solution sets of positive integers a, b, c, d, and e that satisfy a+b+c+d+e = 18.

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The formula for the height above the ground, h(t) is h(t) = 255 sin (π/20 t) + 20.

The amplitude is half the distance between the highest and lowest points, which is (530 - 20)/2 = 255 feet. So A = 255.

The period is 40 minutes, so B = 2π/40 = π/20.

At t = 0 (when we board the Ferris Wheel), we are 20 feet above the ground.

This means there is no phase shift, so C = 0.

The vertical shift is also 20 feet, so D = 20.

Putting it all together, we have:

h(t) = 255 sin (π/20 t) + 20

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The difference of three times a number x and six is greater than or equal to the sum of fifteen and twenty-four times the number is 3x-6≥15+24x

The difference of three times a number and six is greater than or equal to the sum of fifteen and twenty four times a number

Difference is subtraction and sum is nothing but addition

Let the number be x.

The given sentence is changed to the expression or inequality as given below.

3x-6≥15+24x

Hence, the difference of three times a number x and six is greater than or equal to the sum of fifteen and twenty-four times the number is 3x-6≥15+24x

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Answer: B. 6.500

Step-by-step explanation: I just took the quiz.

The equivalent system of first-order differential equations for the given problem is: 1. dv1/dt = v2 2. dv2/dt = v3 3. dv3/dt = 2v2 - v1 + 1 + t*e ^t

Given differential equation: x''' - 2x'' + x' = 1 + t*e ^t

Step 1: Define new variables.
Let's introduce new variables:
v1 = x'
v2 = v1'
v3 = v2'

Now we have:
v1 = x'
v2 = v1'
v3 = v2'

Step 2: Rewrite the given equation using new variables.
Substitute the new variables into the given differential equation:
v3 - 2v2 + v1 = 1 + t*e ^t

Step 3: Write the equivalent system of first-order differential equations.
Now we have the following equivalent system of first-order differential equations:
dv1/dt = v2
dv2/dt = v3
dv3/dt = 2v2 - v1 + 1 + t*e ^t

So, the equivalent system of first-order differential equations for the given problem is:

1. dv1/dt = v2
2. dv2/dt = v3
3. dv3/dt = 2v2 - v1 + 1 + t*e ^t

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Answer: The ratio of A's lateral surface area to B's lateral surface area is 16:1.

Step-by-step explanation: Let B's radius be x and the height be y. Then, the radius of A will be 4x and the height will be 4y.

As we know, the formula for the lateral surface area of a cylinder is

2[tex]\pi[/tex]rh.

So, the lateral surface area of A is 2[tex]\pi[/tex](4x)(4y)= 32[tex]\pi[/tex]xy

lateral surface area of B is 2[tex]\pi[/tex](x)(y)= 2[tex]\pi[/tex]xy

Ratio,

Lateral surface area of A/ Lateral surface area of B = [tex]\frac{32\pi xy}{2\pi xy}[/tex]

=[tex]\frac{16}{1}[/tex]

=16:1

15bags of fiber stuffing will be needed to fill the cones

The volume of one replica ice cream cone is calculated as:

Volume = (1/3) × π × (radius)^2 × height

Radius = 4/2 = 2 inches

Height = 10 inches

Therefore, Volume = (1/3) × π × 2^2 × 10 = 83.78 cubic inches (rounded to two decimal places)

To fill 125 replica cones, the total volume of stuffing required will be:

Total Volume = 125 × 83.78 = 10,472.5 cubic inches

Since each bag of stuffing can fill 720 cubic inches, the number of bags required will be:

Number of bags = 10,472.5 / 720 = 14.55 (rounded up to the nearest whole number)

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The limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

Given: [tex]$\$ P(Z=0)=p \$[/tex], [tex]\$ P(Z=1)=q \$[/tex], [tex]\$ P(Z=3)=r \$$[/tex], where[tex]$\$ p+q+r=1 \$$[/tex] and [tex]$\$ E[Z] < 1 \$$[/tex].

(a) To find the recursion formula for [tex]$\$ W_{-} a(u) \$$[/tex], we use the following formula:

[tex]$$W_a(u)=P(Z=a)+\sum_{k=0}^{u-1} P(Z=a+3 k) W_a(u-1-k)$$[/tex]

where [tex]$\$ \mathrm{a} \$$[/tex] is a non-negative integer and [tex]$\$ \mathrm{u} \$$[/tex] is a positive integer.

Using the given probabilities, we have:

[tex]$$\begin{aligned}& W_0(u)=p+\sum_{k=0}^{u-1} r W_0(u-1-k) \\& W_1(u)=q+\sum_{k=0}^{u-1} p W_1(u-1-k)+\sum_{k=0}^{u-1} r W_1(u-1-k-1) \\& W_3(u)=r+\sum_{k=0}^{u-1} q W_3(u-1-k)\end{aligned}$$[/tex]

(b) To find [tex]$\$ 19$[/tex], we use the fact that [tex]$\$ W_{-} O(u)+W_{-} 1(u)+W_{-} 3(u)=1 \$$[/tex] for all positive integers [tex]$\$[/tex] u [tex]\$$[/tex].

We take the limit as [tex]$\$$[/tex] u [tex]$\$$[/tex] approaches infinity:

[tex]$$\lim _{u \rightarrow \infty} W_0(u)+\lim _{u \rightarrow \infty} W_1(u)+\lim _{u \rightarrow \infty} W_3(u)=1$$[/tex]

Since [tex]$\$ E[Z] < 1 \$$[/tex], we have. Also, from part (a), we have:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\lim _{u \rightarrow \infty} W_0(u-1) \\& \lim _{u \rightarrow \infty} W_1(u)=p \lim _{u \rightarrow \infty} W_1(u-1)+\lim _{u \rightarrow \infty} W_1(u-2)\end{aligned}$$[/tex]

solve for the limits as:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\frac{p}{1-r} \\& \lim _{u \rightarrow \infty} W_1(u)=\frac{q p}{1-p-r}\end{aligned}$$[/tex]

Therefore, we have:

[tex]$$\lim _{u \rightarrow \infty} f(u)=\lim _{u \rightarrow \infty} \frac{W_1(u)}{W_0(u)+W_1(u)}=\frac{q p}{p+(1-r)}=\frac{3}{4}$$[/tex]

Thus, the limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

To learn more about probabilities visit:

https://brainly.com/question/30034780

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