Two teams A and B play a series of games until one team wins four games. We assume that the games are played independently and that the probability that A wins any game is p. What is the probability that the series lasts exactly five games?

Answers

Answer 1
Answer:

Answer:

4pq(p³+q³)

Step-by-step explanation:

Exactly 5 game would be when 4 wins and 1 loss of a particular person

loss has to be one of the first 4 games

A wins: qp⁴ + pqp³ + p²qp² + p³qp

= qp⁴ + qp⁴ + qp⁴ + qp⁴ = 4qp⁴

B wins: pq⁴ + qpq³ + q²pq² + q³pq

= pq⁴ + pq⁴ + pq⁴ + pq⁴ = 4pq⁴

A wins or B wins:

4pq⁴ + 4qp⁴ = 4pq(q³+p³)


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Could someone help me out?
the cube of the sum of 4 and 9 times x divided by the product of 5 times x and the difference of x and 1​
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3x+7x−28+31−8x for x=2043Simplified expression-?Value-?
1. Suppose f(x) = x^4-2x^3+ax^2+x+3. If f(3) = 2, then what is a? 2. Let f, g, and h be polynomials such that h(x) = f(x) * g(x). If the constant term of f(x) is -4 and the constant term of h(x) is 3, what is g(0)? 3. Suppose the polynomials f and g are both monic polynomials. If the sum f(x) + g(x) is also monic, what can we deduce about the degrees of f and g? 4. If f(x) is a polynomial, is f(x^2) also a polynomial 5. Consider the polynomial function g(x) = x^4-3x^2+9 a. What must be true of a polynomial function f(x) if f(x) and f(-x) are the same polynomial b.What must be true of a polynomial function f(x) if f(x) and -f(-x) are the same polynomial

a car travels 27 miles on each gallon of gas. How many miles can the car travel on 8 1/3 gallons of gas

Answers

Basically, we're multiplying 27 by 8 1/3.

1. 27 = 27/1
2. 8 1/3 = 8 * 3 = 24 + 1 = 25/3
3.27/1 * 25/3 = 27 * 25 = 675/3
4.675 divided by 3 = 225

Answer: 225 miles.

Find the surface area of this rectangular prism. Be sure to include the correct unit in your answer.

Answers

Answer: 258 yard^(2)

Step-by-step explanation: To find the surface area you add the area of all the sides. And the unit for a surface area is unit (yards) squared.

Front: 5 • 9 = 45

Back: 5 • 9 = 45

Left: 6 • 5 = 30

Right: 6 • 5 = 30

Top: 9 • 6 = 54

Bottom: 9 • 6 = 54

45 + 45 + 30 + 30 + 54 + 54 = 258.

Final answer:

To calculate the surface area of a rectangular prism, you use the formula 2ab + 2bc + 2ac where a, b, and c represent the length, width, and height of the prism respectively. You compute the surface area in square units.

Explanation:

To find the surface area of a rectangular prism, you need to find the area of all faces and add them up. If the length a, width b, and height c, of the rectangular prism are given, the formula for the surface area is 2ab + 2bc + 2ac square units. Let's take an example: if length (a) is 3 units, width (b) is 4 units, and height (c) is 2 units, then, by using the formula, the surface area would be 2*3*4 + 2*4*2 + 2*3*2 = 24 + 16 + 12 = 52 square units.

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Prove that:
(2-sin(2x))(sin(x) + cos(x)) = 2(sin^3(x) + cos^3(x))

Answers

   
\text{We use formulas: }\n  \n 1) ~~  (a + b)(a^2 -ab + b^2)   =a^3  + b^3 \n  \n 2)~~ \sin(2x) = 2\sin x \cos x  \n \n  3)~~ 1 =\sin^2(x) + cos^2(x) \n  \n \text{We solve:} \n  \n \Big(2-\sin(2x)\Big)\Big(\sin(x) + \cos(x)\Big) = 2\Big(\sin^3(x) + cos^3(x)\Big) \n  \n \Big(2-2\sin(x)\cos(x)\Big)\Big(\sin(x) + \cos(x)\Big) = 2\Big(\sin^3(x) + cos^3(x)\Big) \n  \n 2\Big(1-\sin(x)\cos(x)\Big)\Big(\sin(x) + \cos(x)\Big) = 2\Big(\sin^3(x) + cos^3(x)\Big)


2\Big(\sin^2(x)+\cos^2(x)-\sin(x)\cos(x)\Big)\Big(\sin(x) + \cos(x)\Big) = \n  2\Big(\sin^3(x) + cos^3(x)\Big)  \n  \n 2\Big(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)\Big)\Big(\sin(x) + \cos(x)\Big) = \n  2\Big(\sin^3(x) + cos^3(x)\Big)  \n  \n \boxed{2\Big(\sin^3(x) + cos^3(x)\Big)  = 2\Big(\sin^3(x) + cos^3(x)\Big)  }



John needs to make a scale drawing of his school building for art class. If the building is 256.25 meters long, and John scales it down using a ratio of 25 meters to 1 inch, how long will the building be in the sketch? 12.62 inches

Answers

the answer will be 10.25 inches
256.25 / 25 = 10.25



good luck

What is the next number in the sequence. 1,121,12321, 1234321

The next number in the sequence is _____

Answers

Answer:

123454321

Step-by-step explanation:

it's a palendrome, made out of a number of numbers in the sqquence.

Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. (x2 + y2)y' = y2 1. A unique solution exists in the region y ≥ x.
2. A unique solution exists in the entire xy-plane.
3. A unique solution exists in the region y ≤ x.
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
5. A unique solution exists in the region x2 + y2 < 1.

Answers

A unique solution exists in the region consisting of all points in the xy-plane except the origin.

The correct option is 4.

The given differential equation is:

(x² + y²)y' = y²

The equation can be rewritten as:

x^2 + y^2 (dy)/(dx) = y^2

We need to determine a region of the xy-plane for which the differential equation would have a unique solution whose graph passes through a point (x₀, y₀) in the region.

To determine the region, we can use the existence and uniqueness theorem for first-order differential equations.

According to the theorem, a unique solution exists in a region if the differential equation is continuous and satisfies the Lipschitz condition in that region.

To check if the differential equation satisfies the Lipschitz condition, we can take the partial derivative of the equation with respect to y:

dy/dx = y / (x² + y²)

The partial derivative is continuous and bounded in the entire xy-plane except at the origin (x=0, y=0).

Therefore, the differential equation satisfies the Lipschitz condition in the entire xy-plane except at the origin.

Since the differential equation is continuous in the entire xy-plane, a unique solution exists in any region that does not contain the origin. Therefore, the correct answer is:

A unique solution exists in the region consisting of all points in the xy-plane except the origin.

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Final answer:

The differential equation will have a unique solution in the entire xy-plane except at the origin, as both the function and its partial derivatives are continuous and well-defined everywhere except at that point.

Explanation:

To determine a region of the xy-plane where the differential equation (x2 + y2)y' = y2 has a unique solution passing through a point (x0, y0), we need to consider where the function and its derivative are continuous and well-defined. According to the existence and uniqueness theorem for differential equations, a necessary condition for a unique solution to exist is that the functions of x and y in the equation, as well as their partial derivatives with respect to y, should be continuous in the region around the point (x0, y0).

We note that both the function (x2 + y2)y' and its partial derivative with respect to y, which is 2y, are continuous and well-defined everywhere except at the origin where x = 0 and y = 0. Therefore, a unique solution exists in the region consisting of all points in the xy-plane except the origin.

From the given options, the correct answer is:

4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.

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