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3 years ago

O COUNTING AND PROBABILITY

Mathematics

1 answer:

valina [46]3 years ago

4 0

Answer:

a. $Odds = \frac{3}{7}$

b. $Odds = \frac{7}{3}$

Step-by-step explanation:

Given

$n(Black) = 7$

$n(Grey) = 2$

$n(White) = 1$

$n(Total) = 10$

Solving (a): Odds against winning

Since, the arrow must stop at black, then

The odds against winning is calculated as thus:

$Odds = \frac{n(Total) - n(Black)}{n(Black)}$

$Odds = \frac{10 - 7}{7}$

$Odds = \frac{3}{7}$

Solving (b): Odds in favor of winning

Since, the arrow must stop at black, then

The odds is calculated as thus:

$Odds = \frac{n(Black)}{n(Total) - n(Black)}$

$Odds = \frac{7}{10 - 7}$

$Odds = \frac{7}{3}$

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Answer:

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6 cups of rice

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When asking how many this will serve, we also multiply by 3 the original amount of people which is 4 to get 12 servings of Arroz con leche

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3 years ago

Enter the fraction as a decimal

Andrew [12]

$\frac { 0.2 } { 5 } \\ = \frac{0.2 \times 10}{5 \times 10} \\ = \frac{2}{50} \\ = \frac{1}{25} \\ = \boxed{0.04}$

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3 years ago

Read 2 more answers

Pls help asap thanks .

lys-0071 [83]

Answer:

n/a

Step-by-step explanation:

A. t(h) = 5.5h + 50

B. t(6) = 5.5(6) + 50

t(6) = 85

C. (6, 85) This means that when the electrician for 7 hours, they get paid $85

8 0

3 years ago

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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

Need help 10 points

Citrus2011 [14]

Answer:

50.24

Step-by-step explanation:

8 is the radius of the circle, so using A=πr^2, the equation would be A=(3.14*8^2)/4

(Dividing by 4 because we are solving for 1/4 of a circle, and substituting pi for 3.14)

(3.14*8^2)=200.96

200.96/4=50.24

3 0

2 years ago