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butalik [34]
3 years ago
10

The probability that the Cubs win their first game is 1/3. The probability that the Cubs win their second game is 3/7. What is t

he probability that the Cubs win games?
Mathematics
2 answers:
olga55 [171]3 years ago
6 0
We need to multiply these probabilities up. Indeed, we are talking about independent events. Then, the final answer is 1/3 * 3/7=1/7
vaieri [72.5K]3 years ago
6 0

The <em>correct answer</em> is:


1/7.


Explanation:


The two events, the Cubs winning their first game and the Cubs winning their second game, are independent. This is because the Cubs winning their first game does not affect the probability of the Cubs winning their second game.


Since the events are independent, to find the probability that both events occur, multiply the probabilities:


1/3(3/7) = (1*3)/(3*7) = 3/21 = 1/7

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Read 2 more answers
A quadratic function is shown. f(x) = -x^2+7x-12 one factor of this function is (x-4). What is the other factor of the quadratic
Alexus [3.1K]

Answer:

The other term is<u> (x-3)</u>

Step-by-step explanation:

The quadratic function is:

f(x)=-x^{2}+7x-12

Factoring this function we will have:

f(x)=-(x-4)(x-3)

Therefore, the other term is<u> (x-3).</u>

I hope it helps you!

3 0
3 years ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
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