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3 years ago

Players a, b, and c toss a fair coin in order. the first to throw a head wins. what are their respective chances of winning

1 answer:

8 0

Probability a gets a head = 1/2
.. .. .. b .. .. ... .. = P(a gets tail) * 1/2 = 1/2 * 1/2 = 1/4
.. ... . .. c .. .. .. .. . = P(and and b gets tail) * 1/2 = 1/4 * 1/2 = 1/8

You might be interested in

Please help with part b

timofeeve [1]

Answer:

161391

Step-by-step explanation:

138000 x 16.95/100 = 23391

138000 + 23391 = 161391

5 0

3 years ago

Can someone please tell me how to find the missing length of 27 and 28?

denis23 [38]

The square root of 117

6 0

3 years ago

Which expression has the same value as 37.4 - 54.9?

GrogVix [38]

Answer:

Step-by-step explanation:

In the original expression = 37.4 - 54.9 = -17.5

So, we have to solve all the equations but 1 of them has to equal to -17.5

A) -54.9 + (-37.4) = -92.3

Remove parentheses

= -54.9 - 37.4

Subtract the numbers

-92.3

B) 54.9 - 37.4 = 17.5

Since 54.9 is greater than 37.4 just subtract normally

4 14

5 4. 9 < ------- Line up the decimal points like this

- 3 7. 4 <--------- Subtract the decimals

-----------------

1 7 . 5 <---------- This is the final answer to B.

C) 37.4 - (- 54.9) = 92.3

Change the sign to addition

=37.4 + 54.9

Add the numbers normally

54.9

+ 37. 4

--------------

92.3

D) 37.4 + (-54.9) = -17.5

Change the sign to subtraction

= 37.4 - 54.9

Subtract them normally (but as you can see you are subtracting a smaller number from a larger number so heres a trick that you can use switch the equation by putting the larger number in first like this....)

54.9 - 37.4 = 17.5 (REMEMBER the answer is not 17.5 because the original equation of D is reversed 37.4 - 54.9 so in this case your answer should be

- 17.5)

37.4 - 54.9 = -17.5

Your final answer is D because it matches 37.4 - 54.9.

Hope this helps!

3 0

3 years ago

Read 2 more answers

How do I find the percentage of 18/20

borishaifa [10]

Or for this and any other problem . every percentage is out of a 100

soooo ,
20 | 100 %
10 | 50 %
5 | 25 %
1 | 5 %

SOOO

10 + 5 = 15 50 % + 25 % = 75 % , so know we know 15/20 is 75 percent

but you need 18/20 not 15/20 , so 1 equals 5 % in this problem

1+1+1 = 3 5+5+5 = 15

15 + 3 = 90 75 + 15 = 90 , 18/20 = 90%

I know its probably alot but I wanted to show you all I did , so that it could maybe help you in the future , just gotta plug in different numbers

4 0

4 years ago

Read 2 more answers

Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d

Leni [432]

Answers:

(a) $y = \frac{1}{1 - Cx}$ , for any constant C

(b) Solution does not exist

(c) $y = \frac{256}{256 - 15x}$

(d) $y = \frac{64}{64 - 15x}$

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

$x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}$ (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

$\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }$ (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

$1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)} \end{cases}$

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1

Hence,

$\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}$

So,

$\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}$

Now, equation (1) becomes

$\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent 1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
$ (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

$

Hence, for any constant C, the following solution will pass thru (0, 1):

$\boxed{y = \frac{1}{1 - Cx}}$

(b) Using equation (5) in problem (a),

$y = \frac{1}{1 - Cx}$ (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that

$y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1$

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C.

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}


$

By replacing this value of C, the general solution becomes

$y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}



$

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
- Substitute the values of x and y to the general solution.
- Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that

$y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}$

Now, we replace C using the derived value in the general solution. Then,

$y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}$

5 0

3 years ago