find an equation of the curve that passes through the point (0, 1) and whose slope at (x, y) is 13xy.

This means that we can be 95% confident that the true difference in means between Clarion and Wabash is somewhere between -3.33 and -0.88

A 95% confidence interval for the difference in means between Clarion and Wabash can be calculated using the following formula: CI95 = (μ1 - μ2) ± 1.96*√(σ1^2/n1 + σ2^2/n2),where μ1 and μ2 are the population means of Clarion and Wabash respectively, σ1 and σ2 are the population standard deviations of Clarion and Wabash respectively, and n1 and n2 are the sample sizes of Clarion and Wabash respectively. To calculate the confidence interval, we need to have access to the population means and standard deviations of Clarion and Wabash, which we do not have. In their place, we can use the sample means and standard deviations as an estimate of the population means and standard deviations. Using the sample means and standard deviations, the 95% confidence interval for the difference in means between Clarion and Wabash is (-3.33, -0.88). This means that we can be 95% confident that the true difference in means between Clarion and Wabash is somewhere between -3.33 and -0.88.

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What is the approximate 95% confidence interval for the difference in means between Clarion and Wabash?

The expected amount of time until the ant returns to its starting vertex is 2.5 minutes.

Vertex in math is the point at which two lines, curves, or edges meet. In the case of a triangle, for example, the vertex is the point where all three lines meet. In higher dimensional shapes, such as a parallelogram, it is the point at which all the lines meet. Vertex can also be used to refer to the highest point of a graph, or the point of maximum or minimum value.

The expected amount of time until an ant returns to its starting vertex after traversing the edges of a tetrahedron can be calculated by applying the principle of expected value. The expected value of a random variable is the sum of the probability of each outcome multiplied by its associated value.
In this case, the ant has four possible outcomes, with each outcome being the length of time it takes to traverse the edge to an adjacent vertex. Since the ant has an equal probability of going to each vertex, each outcome has a probability of 0.25. Thus, the expected value can be calculated as:
Expected Value = (1 minute x 0.25) + (2 minutes x 0.25) + (3 minutes x 0.25) + (4 minutes x 0.25)
Expected Value = 2.5 minutes
Therefore, the expected amount of time until the ant returns to its starting vertex is 2.5 minutes.

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Complete questions as follows-
an ant starts at one vertex of a tetrahedron. each minute it walks along a random edge to an adjacent vertex. what is the expected amount of time until it returns to its starting vertex?

We can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability.

What is probability?

Probability is simply how likely something is to happen. Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

The theoretical probability of rolling a 5 on a fair die is 1/6, which means that if the die is rolled many times, we would expect to see a 5 about 1/6 of the time.

For the first 100 trials, Maya rolled a 5 on 25 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 25/100

experimental probability = 0.25

So, in the first 100 trials, Maya's experimental probability of rolling a 5 was 0.25.

For the first 200 trials, Maya rolled a 5 on 30 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 30/200

experimental probability = 0.15

So, in the first 200 trials, Maya's experimental probability of rolling a 5 was 0.15.

Comparing these experimental probabilities to the theoretical probability, we see that after 100 trials, Maya's experimental probability of rolling a 5 (0.25) is higher than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 100 trials was somewhat biased in favor of rolling a 5.

On the other hand, after 200 trials, Maya's experimental probability of rolling a 5 (0.15) is lower than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 200 trials was somewhat biased against rolling a 5.

Overall, we can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability. This is known as the law of large numbers, which states that as the number of trials or observations increases, the experimental probability will tend to approach the theoretical probability.

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We might say that Maya's experimental probabilities oscillate about the theoretical probability, but after more trials, the experimental probabilities ought to converge to the theoretical probability.

What is probability?

Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

A fair die has a theoretical probability of rolling a 5 of 1/6, therefore if the die is rolled several times, we can anticipate seeing a 5 roughly 1/6 of the time.

For the first 100 trials, Maya rolled a 5 on 25 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 25/100

experimental probability = 0.25

So, in the first 100 trials, Maya's experimental probability of rolling a 5 was 0.25.

For the first 200 trials, Maya rolled a 5 on 30 of those trials. The experimental probability of rolling a 5 in this case is:

experimental probability = number of 5's rolled / number of trials

experimental probability = 30/200

experimental probability = 0.15

So, in the first 200 trials, Maya's experimental probability of rolling a 5 was 0.15.

Comparing these experimental probabilities to the theoretical probability, we see that after 100 trials, Maya's experimental probability of rolling a 5 (0.25) is higher than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 100 trials was somewhat biased in favor of rolling a 5.

On the other hand, after 200 trials, Maya's experimental probability of rolling a 5 (0.15) is lower than the theoretical probability (1/6 ≈ 0.167). This suggests that Maya's sample of 200 trials was somewhat biased against rolling a 5.

Overall, we can conclude that Maya's experimental probabilities fluctuate around the theoretical probability, but over a larger number of trials, the experimental probabilities should converge towards the theoretical probability. This is known as the law of large numbers, which states that as the number of trials or observations increases, the experimental probability will tend to approach the theoretical probability.

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Therefore , the solution of the given problem of unitary method comes out to be common shares of Zeil Inc. is 2.23%.

This common convenience, already-existing variables, or all important elements from the original Diocesan adaptable study that followed a particular methodology can all be used to achieve the goal. Both of the crucial elements of a term affirmation outcome will surely be missed if it doesn't happen, but if it does, there will be another chance to get in touch with the entity.

Here,

Earnings per Share are calculated as (Net Income – Preferred Dividends) / the average number of outstanding Common Shares.

=>  Market price per share / earnings per share is the Price-Earnings Ratio.

=> Dividends per Share are calculated as follows: Common Stock Dividends / Average Common Shares Outstanding

=>  Dividend Yield is the product of dividends per share and the share price.

=>  (Beginning Common Shares plus Ending Common Shares) / 2 equals the average number of Common Shares Outstanding.

=>  Starting common shares equals ending common shares, which is

=>  $3,500,000 / $25, or 140,000.

(a) The earnings per share are ($424,000 - $0) / 140,000, which equals $3.03.

The ordinary stock price of Zeil Inc.

(b) The price-earnings ratio for Zeil Inc.'s common shares is 11.20 divided by 3.03, or 3.69.

(c) Dividends per Share: $35,000./140,000. = $0.25

Therefore, $0.25 in dividends are paid per unit of Zeil Inc. common stock.

(d) Dividend Yield: $0.25 divided by $11.20 equals 0.0223, or 2.23%.

The common shares of Zeil Inc. is 2.23%.

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The probability that the sample mean of ln(LC50) will differ from the true population mean by no more than 0.5 is 0.0019.

The concept of independent random variables is very similar to independent events. Recall that two events A and B are independent if we have P(A,B) = P(A)P(B) and remember that comma means sum, i.e.

         P(A, B)=P( A and B)=P (A∩B).

Similarly, we have the following definition for independent discrete random variables.

Assuming 'X' and 'Y' are independent random samples

X : mean value =  , variance = 0.4 /10 = 0.04

Y : mean value =  , variance = 0.8/10 = 0.08

since the values are independent

V [ X - Y ] = V [ X ] + V [Y ] = 0.04 + 0.08 = 0.12

Now,

The probability that the sample mean for species A exceeds the sample mean for species B by at least 1 unit

  [tex]P(X-Y\geq 1) = P{ \frac{(X-Y)-(U_1-U_2)}{\sqrt{V(X-Y} } \geq \frac{1-0}{\sqrt{0.12} }[/tex]

⇒ P{Z ≥ 2.8858}

⇒ 1 -P{Z≤ 2.8858} = 1 - 0.9981

⇒ P{X-Y≥ 1} = 0.0019.

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Yes, there is an analogous ratio between 2:1 and 20:10.

We just cancel by a common factor. So 4:2=2:1 . The simplest representation of the ratio 4 to 2 is the ratio 2 to 1. Also, since each pair of numbers has the same relationship to one another, the ratios are equivalent.

By dividing the terms of each ratio by their greatest common factor, we may simplify both ratios to explain why.

As the greatest common factor for the ratio 2:1 is 1, additional simplification is not necessary.

The greatest common factor for the ratio 20:10 is 10. When we multiply both terms by 10, we get:

20 ÷ 10 : 10 ÷ 10

= 2 : 1

As a result, both ratios have the same reduced form, 2:1, making them equal.

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Therefore, the monthly payment for a student loan of $15,000 at a fixed APR of 12% for 20 years is $144.36.

Interest is the cost of borrowing money or the return on investing money. When you borrow money, you usually have to pay back more than you borrowed, and the additional amount you pay is the interest. The interest rate is expressed as a percentage of the borrowed amount, and it can vary depending on factors such as the borrower's credit score, the term of the loan, and the lender's policies.

Given by the question.

Assuming the loan has a fixed interest rate of 12% per annum, the amount of interest charged each year will be:

12% of $15,000 = $1,800

The total interest charged over 20 years will be:

$1,800 x 20 = $36,000

The total amount to be repaid (principal + interest) will be:

$15,000 + $36,000 = $51,000

If the loan is being repaid in equal monthly installments over the 20-year term, the monthly payment can be calculated using the following formula:

M = P * (r[tex](1+r)^{n}[/tex]) / ([tex](1+r)^{n}[/tex]- 1)

Where:

M = Monthly payment

P = Principal amount (in this case, $15,000)

r = Monthly interest rate (12% per annum / 12 months = 1% per month)

n = Total number of payments (20 years x 12 months per year = 240)

Plugging in the values:

M = $15,000 * (0.01[tex](1+0.01)^{240}[/tex]) / ([tex](1+0.01)^{240}[/tex] - 1)

M = $144.36

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The given statement " show that there exist coefficients w0, w1, ..., wn that depend on x0, x1, ..., xn, and on a and b, such that the limit of the sum, as a approaches b, of the form summation from i=0 to n of wi*p(xi) for all polynomials p of degree <= n", is proved by the use of Lagrange form of the interpolating polynomials.

Let p(x) be a polynomial of degree at most n. Then, by the Lagrange interpolation formula from Section 4.1, we have:

p(x) = Summation from i=0 to n of p(xi) * Li(x)

where Li(x) is the ith Lagrange basis polynomial, defined by:

Li(x) = Product from j=0 to n, j != i, of (x - xj) / (xi - xj)

Now, consider the sum:

S = Summation from i=0 to n of wi * p(xi)

where wi are coefficients to be determined. We want to show that the limit of S as a approaches b exists for all polynomials p of degree at most n.

We can express S in terms of the Lagrange basis polynomials as:

S = Summation from i=0 to n of wi * p(xi)

= Summation from i=0 to n of wi * Summation from j=0 to n of p(xj) * Li(xj)

= Summation from j=0 to n of p(xj) * Summation from i=0 to n of wi * Li(xj)

Note that the summation over i is only dependent on the Lagrange basis polynomial Li(xj), and does not depend on p(xj). Therefore, we can choose the coefficients wi such that:

Summation from i=0 to n of wi * Li(xj) = 0 for j != k

Summation from i=0 to n of wi * Li(xk) = 1

for some k in {0, 1, ..., n}.

To see why this is possible, note that the Lagrange basis polynomials satisfy the property that Li(xi) = 1 and Li(xj) = 0 for j != i. Therefore, we can choose the coefficients wi to be:

wi = Li(xk) / Summation from i=0 to n of Li(xk)

which gives:

Summation from i=0 to n of wi * Li(xj) = Li(xk) / Summation from i=0 to n of Li(xk) * Summation from i=0 to n, i != k of Li(xj)

= 0 for j != k

Summation from i=0 to n of wi * Li(xk) = 1

Now, we have:

S = Summation from j=0 to n of p(xj) * Summation from i=0 to n of wi * Li(xj)

= Summation from j=0 to n of p(xj) * Li(xk)

Taking the limit as a approaches b, we get:

lim a->b S = lim a->b Summation from j=0 to n of p(xj) * Li(xk)

= Summation from j=0 to n of p(xj) * lim a->b Li(xk)

= Summation from j=0 to n of p(xj) * Integral from a to b of Li(x) dx

where we have used the fact that the limit and integral commute, and the limit of the Lagrange basis polynomial Li(xk) is equal to the integral of Li(x) over the interval [a, b], which is a constant that does not depend on k.

Therefore, we have shown that there exist coefficients w0, w1, ..., wn that depend on x0, x1, ..., xn, and on a and b, such that the limit of the sum, as a approaches b, of the form Summation from n to i=0 wi p(xi).

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_____The given question is incomplete, the complete question is given below:

Show that there exist coefficients w0,w1, . . . ,wn depending on x0, x1, . . . , xn and on a, b such that limit a to b { summation n to i=0 wi p(xi)} for all polynomials p of degree ?n.

   Hint: Use the Lagrange form of the interpolating polynomials from Section 4.1

Answer:

To find the probability that a randomly selected person is a meat-eater, we need to add up the number of meat-eaters and divide by the total number of individuals surveyed. From the given table, we can see that there are 72 meat-eaters out of a total of 190 individuals surveyed:

Total meat-eater = 72

Total surveyed = 190

So the probability of selecting a meat-eater is:

P(meat-eater) = Total meat-eater / Total surveyed

P(meat-eater) = 72 / 190

P(meat-eater) = 0.38 (rounded to the hundredths place)

Therefore, the probability that a randomly selected person is a meat-eater is 0.38 or 38%.

Toward the end of a game of Scrabble, you hold the letters D, O, G, and Q. You can choose 3 of these 4 letters and arrange them in order in 24 different ways.

To solve this problem, we need to use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we need to find the number of permutations that can be made from the letters D, O, G, and Q when we choose 3 of these 4 letters.

The formula for finding the number of permutations is:

n! / (n-r)!

where n is the total number of objects and r is the number of objects we choose.

Using this formula, we can calculate the number of permutations as follows:

4! / (4-3)!

= 4! / 1!

= 4 x 3 x 2 x 1 / 1

= 24

Therefore, we can arrange the chosen 3 letters in 24 different ways.

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Triangle pair (A, B) can be mapped to each other using a translation and rotation about point A, which is calculated using the formula T(x, y) = (x + a, y + b) followed by a rotation of θ about the point A(a, b).

The triangle pair (A, B) can be mapped to each other using a translation and rotation about point A. The formula for the transformation is T(x, y) = (x + a, y + b) followed by a rotation of θ about the point A(a, b). The calculation for this transformation would be x' = xcosθ - ysinθ + a and y' = xsinθ + ycosθ + b. For example, if the coordinates of point A are (4, 5), point B is (6, 2) and the rotation angle is 30°, the coordinates of point B' would be (7.9, 5.4) after the transformation.

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Answer:

We can only see g(x) not f(x)

Step-by-step explanation:

Domain of g(x) is

[tex]( - \infty \: to \: \infty )[/tex]

Range of g(x) is

[tex](0 \: to \: \infty )[/tex]

Range lf

The slope of this linear function is equal to: B. -2/9.

The volume of a cylinder with a height of 10 m and a radius of 5 m is equal to 785 m³.

The value of each expression is: C. a) 2, b) 1/2, c) 2/9.

In Mathematics, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (8 - 10)/(6 - (-3))

Slope (m) = (8 - 10)/(6 + 3)

Slope (m) =

Slope (m) = -2/9.

In Mathematics, the volume of a cylinder can be calculated by using this formula:

Volume of a cylinder, V = πr²h

Where:

By substituting the given parameters, we have:

Volume of cylinder, V = 3.14 × 5² × 10

Volume of cylinder, V = 785 m³

(√2)² = 2

(1/√2)² = 1/2

(√2/3)² = 2/9

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Answer:

Step-by-step explanation:

To prove that z^2 + 1/2z is a real number, we need to show that the imaginary part of z^2 + 1/2z is equal to zero.

We know that z = (a+bi)/(a-bi)

Multiplying the numerator and denominator by the complex conjugate of the denominator, we get

z = (a+bi)(a+bi)/(a-bi)(a+bi)

z = (a^2 + 2abi - b^2)/(a^2 + b^2)

Expanding z^2, we get:

z^2 = [(a^2 + 2abi - b^2)/(a^2 + b^2)]^2

z^2 = (a^4 + 2a^2b^2 + b^4 - 2a^2b^2 + 4a^2bi - 4b^2i)/(a^4 + 2a^2b^2 + b^4)

Simplifying, we get:

z^2 = (a^4 - b^4 + 2a^2bi)/(a^4 + 2a^2b^2 + b^4)

Now, let's compute z^2 + 1/2z:

z^2 + 1/2z = (a^4 - b^4 + 2a^2bi)/(a^4 + 2a^2b^2 + b^4) + 1/2[(a+bi)/(a-bi)]

To simplify this expression, we need to find a common denominator:

z^2 + 1/2z = (2a^5 - 2a^3b^2 + 3a^4b - 3ab^4 - 2b^5 + 3a^3bi + 3ab^3i)/(2(a^4 + 2a^2b^2 + b^4))

We can see that the imaginary part of z^2 + 1/2z is (3a^3b - 3ab^3)/(2(a^4 + 2a^2b^2 + b^4))

However, we know that a and b are real numbers, so the imaginary part of z^2 + 1/2z is zero.

Therefore, z^2 + 1/2z is a real number.

To calculate the molarity of Pb(NO3)2(aq), you need to know the molar mass of the compound. The molar mass of Pb(NO3)2 is 331.21 g/mol.

Using the equation, concentration = moles/liters, we can calculate the molarity of the Pb(NO3)2(aq) solution.

First, we need to calculate the moles of Pb(NO3)2. We can do this by converting the mass of the precipitate (19.40 g) to moles. Moles = mass (g) / molar mass (g/mol).

Therefore, moles of PbCl2 = 19.40 g / 331.21 g/mol = 0.05833 moles.

Next, we can calculate the molarity of Pb(NO3)2. Molarity = moles/liters.

Therefore, the molarity of Pb(NO3)2 = 0.05833 moles/ 0.2 liters = 0.29165 M.

Answer:

19/20 , 1 1/4 , 1 4/5

Step-by-step explanation:

1.8 = 1 4/5 (fraction)

1.8 converts to 18/10. This can be simplified twice, firstly by making it 9/5 since both 18 and 10 are divisible by two, but can be simplified further to 1 4/5

1.25 = 1 1/4 (fraction)

1.25 converts to 125/100. This can be simplified to 5/4 or 1 1/4

0.95 = 19/20 (fraction)

0.95 converts to 95/100. This can be simplified to 19/20

Ascending Order (smallest to largest)

smallest - 19/20

middle - 1 1/4

largest - 1 4/5

I believe this is the right answer, but haven't done fractions in a while so may want to double check to make sure

Step-by-step explanation:

Please mark as brainliest

Answer:

36.3636364%

or 36.36

Step-by-step explanation:

Jadi, lebar jalan adalah 14 meter.

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So Abbie was asked to randomly select at least 368 of her high school unitary method students to estimate the percentage of students planning to go to college. With a 90% confidence level and a 5% error rate. 

The unit method is an approach to problem solving that first determines the value of a single unit and then multiplies that value to determine the required value. Simply put, the unit method is used to extract a single unit value from a given multiple. For example, 40 pens cost 400 rupees or pen price. This process can be standardized. single country. Something that has an identity element. (Mathematics, Algebra) (Linear Algebra, Mathematical Analysis, Matrix or Operator Mathematics) Adjoints and reciprocals are equivalent. 

To determine the required sample size, the following formula should be used:

[tex]n = (Z^2 * p * (1-p)) / E^2[/tex]

where:

N:sample size required

Z:The Z-score corresponds to the desired confidence level and is 1.645 at the 90% confidence level.

Pa:Estimated Percentage of Students Planning to Go to College

1-p:Percentage of students not planning to go to college

E:Desired error margin of 0.05

Since we don't know the actual percentage of students who want to go on to college, we must use estimates based on past studies and surveys. Let's assume the estimated proportion is 0.6 (her 60% of students).

After plugging in the values ​​it looks like this:

[tex]n = (1.645^2 * 0.6 * 0.4) / 0.05^2\\n = 368.03[/tex]

So Abbie was asked to randomly select at least 368 of her high school students to estimate the percentage of students planning to go to college. With a 90% confidence level and a 5% error rate. 

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Answer:

Let's work backwards from the end of lesson 4 to figure out how many sweets were left in Anna's bag after each lesson. We know that she had 1 sweet left at the end of lesson 4, so before that she must have had:

Lesson 4: 1 sweet + 1 sweet for teacher + 1 sweet left over = 3 sweets

Lesson 3: (3 sweets + 1 sweet for teacher) x 2 = 8 sweets

Lesson 2: (8 sweets + 1 sweet for teacher) x 2 = 18 sweets

Lesson 1: (18 sweets + 1 sweet for teacher) x 2 = 40 sweets

So, Anna started with 40 sweets in her bag.

If Andres Michael bought a new boat. He took out a loan for $24,420 at 3.5% interest for 2 years. Andres Michael owes $18806.6 at maturity.

To calculate how much is due at maturity, we first need to determine how much of the loan remains after the two partial payments.

To do this, we can use the formula for simple interest:

I = P * r * t

Where:

I = Interest

P = Principal (original loan amount)

r = Annual interest rate

t = Time (in years)

The interest for the first two months can be calculated as:

I1 = P * r * t1

= 24420 * 0.035 * (2/12)

= 142.45

So after the first two months, the amount owing on the loan is:

P1 = P + I1 - 4330

= 24420 +142.45 - 4330

= 20,232.45

The interest for the next four months can be calculated as:

I2 = P1 * r * t2

= 20,232.45 * 0.035 * (4/12)

= 236.05

So after six months, the amount owing on the loan is:

P2 = P1 + I2 - 2600

=  20,232.45 + 236.05- 2600

= 17868.50

Now we can calculate the interest for the remaining 18 months:

I3 = P2 * r * t3

=  17868.50* 0.035 * (18/12)

= 938.10

So the total amount owing at maturity (after 2 years) is:

Total amount owing = P2 + I3

=  17868.50 + 938.10

= 18806.6

Therefore, Andres Michael owes $18806.6 at maturity.

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The joint probability density function for the random variables Y1 and Y2 E(Y1+Y2) and V(Y1+Y2) is 41/144.

To find E(Y1+Y2), we need to integrate the sum of Y1 and Y2 over their joint probability density function:

E(Y1+Y2) = ∫∫ (y1 + y2) f(y1,y2) dy1 dy2

= ∫∫ (y1 + y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 <=1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex](y1 + y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (2y1 + 1) (1-y1)² dy1

= 5/12

To find V(Y1+Y2), we can use the formula V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]².

First, we need to find E[(Y1+Y2)^2]:

E[(Y1+Y2)²] = ∫∫ (y1+y2)² f(y1,y2) dy1 dy2

= ∫∫ (y1² + y2² + 2y1y2) (2) dy1 dy2, where the limits of integration are 0 to 1 for both y1 and y2 and y1+y2 = 1

= ∫[tex]0^1[/tex] ∫[tex]0^{(1-y1)}[/tex] (y1² + y2² + 2y1y2) (2) dy2 dy1

= ∫[tex]0^1[/tex] (1/3)y1³ + (1/2)y1² + (1/2)y1

(1/3)y1 + (1/4) dy1

= 7/12

Next, we need to find [E(Y1+Y2)]²:

[E(Y1+Y2)]² = (5/12)² = 25/144

Therefore, V(Y1+Y2) = E[(Y1+Y2)²] - [E(Y1+Y2)]² = (7/12) - (25/144) = 41/144.

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Answer:

$586.00

Step-by-step explanation:

Markup is the how much more an item or service is sold for to cover overhead fees. If the markup is 25%, then the price was increased by 25% in order to be sold for $732.50. We can set up a proportion to represent this where c is the cost.

[tex]\frac{732.5}{1.25} = \frac{c}{1.00}[/tex]

Cross-multiply.

1.25c = 732.5

c = 586

So, the cost of the item was $586.00

If x and y are integer variables, then the expression that evaluates to true if x is greater than y is "x>y".

In Java, symbol of ">" is used for "greater-than" operator. So, the expression which evaluates to "true" if integer "x" is greater than integer "y" is "x > y".

This expression compares the values of x and y and returns a Boolean value of "true" if x is greater than y, and "false" otherwise.

The expression can be used in conditional statements, loops, and other constructs that require a Boolean value as a condition. It is important to note that the ">" operator only works with primitive types such as int, long, double, etc.

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By answering the question the answer is standard deviation  For 0 courses: 3 students (3/55 ≈ 0.1 or 9.1%); For 1 course: 12 students (12/55 ≈ 0.2 or 21.8%)

Standard deviation is a statistic that describes the variability or variance of a group of numbers. A high standard deviation indicates that the values ​​are more dispersed, while a low standard deviation indicates that the values ​​tend to be closer to the established mean. A measure of how far the data are from the mean is the standard deviation (or ). If the standard deviation is small, the data tend to be clustered around the mean, and if the standard deviation is large, the data are more dispersed. The average variability of the dataset is measured as standard deviation. Shows the mean deviation of each score from the mean. 

To fill in the blank cells, we need to calculate the number of students who reported each course number and the relative frequency (rounded to the nearest tenth). This can be done like this:

For 0 courses:

3 students (3/55 ≈ 0.1 or 9.1%)

For 1 course:

12 students (12/55 ≈ 0.2 or 21.8%)

For 2 courses:

17 students (17/55 ≈ 0.3 or 30.9%)

For 3 courses:

9 students (9/55 ≈ 0.2 or 16.4%)

For 4 courses:

8 students (8/55 ≈ 0.1 or 14.5%)

For 5 courses:

2 students (2/55 ≈ 0.0 or 3.6%)

For 6 courses:

4 students (4/55 ≈ 0.1 or 7.3%)

The finished table looks like this:

+--------+--------+---------------------+

| Number | Number | Relative Frequency  |

|  of    | of     | (Rounded to nearest |

|Courses |Students|         tenth)      |

+--------+--------+---------------------+

|   0    |   3    |         0.1         |

|   1    |   12   |         0.2         |

|   2    |   17   |         0.3         |

|   3    |   9    |         0.2         |

|   4    |   8    |         0.1         |

|   5    |   2    |         0.0         |

|   6    |   4    |         0.1         |

+--------+--------+---------------------+

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If a₁ = 5 and an 5an-1 then The value οf a₄ is 625.

An arithmetic sequence is a sequence οf numbers in which each term after the first is fοund by adding a fixed cοnstant number, called the cοmmοn difference, tο the preceding term. Fοr example, the sequence 2, 5, 8, 11, 14, ... is an arithmetic sequence with a cοmmοn difference οf 3, since each term after the first is fοund by adding 3 tο the preceding term.

The nth term οf an arithmetic sequence can be fοund using the fοrmula:

an = a1 + (n-1)d

where an is the nth term, a1 is the first term, and d is the cοmmοn difference. The sum οf the first n terms οf an arithmetic sequence can be fοund using the fοrmula:

Sn = n/2 (a1 + an)

We are given that a₁ = 5, and that the nth term is 5 times the (n-1)th term. We can use this infοrmatiοn tο find the value οf a₄ as fοllοws:

a₂ = 5a₁ = 5(5) = 25

a₃ = 5a₂ = 5(25) = 125

a₄ = 5a₃ = 5(125) = 625

Therefore, the value of a₄ is 625.

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Based on the differential equation and the initial condition, the population of the mouse species never becomes zero. Therefore, there is no time at which the population becomes zero.

We can begin by separating variables and integrating both sides of the equation

dt/dp(t) = 21/p(t) - 450

dt = (1/21) * (1/p(t) - 450) dp(t)

Integrating both sides gives

t + C = (1/21) * ln|p(t)| + 450t + D

where C and D are constants of integration. We can solve for these constants using the initial condition p(0) = 850

0 + C = (1/21) * ln|850| + 0 + D

C = (1/21) * ln|850| - D

We can simplify this expression by defining a new constant E = (1/21) * ln|850| - D

C = E - D

Substituting this expression for C back into our previous equation, we have

t + E - D = (1/21) * ln|p(t)| + 450t

Solving for p(t), we get

ln|p(t)| = 21(450t + D - E) + ln|850|

p(t) = ± e^(21(450t + D - E) + ln|850|)

Since p(t) represents a population, we can discard the negative solution and take only the positive solution

p(t) = e^(21(450t + D - E) + ln|850|)

We want to find the time at which the population becomes zero, so we set p(t) = 0 and solve for t

0 = e^(21(450t + D - E) + ln|850|)

ln|0| = 21(450t + D - E) + ln|850|

This is not possible, since ln|0| is undefined. Therefore, the population never becomes zero.

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