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astra-53 [7]
4 years ago
10

For a given a event what is the result of dividing the number of successful outcomes by the number of possible outcomes

Mathematics
1 answer:
galina1969 [7]4 years ago
8 0
The number of successful outcomes can be represented by p while number of unsuccessful outcomes can be represented by q. In this case, the sum of p and q is equal to the sample space, regarded here as n. so, taking the quotient of the two:

p/q = p / n-q 

If these were expressed in probabilities:
p/q = p / 1- p

The quotient would be greater than 1.
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Y = -2x +8
y = x-1

y=y

-2x+8 = x-1

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-3x = -9

3x = 9

x = 3


 
6 0
3 years ago
A road crew is widening a street that is 24 m wide. Their scale drawing says the new street has to be 125% of the width of the o
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Answer: 30 m

Step-by-step explanation: i think this is correct and i apologize if it’s not, but there’s two ways you could do this. 1: initial X increase/decrease in decimal form. in this case, 24 X 1.25 (125% in decimal form). 2: you could figure out what 25% of 24 meters is in the same way as method 1, and then multiply that answer by five to get 125%. i reallyyy hope this helped and if it’s wrong i’m really sorry, but good luck!

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4 years ago
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ad-work [718]

Answer:

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Step-by-step explanation:

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3 years ago
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Given that y = sin(x+y),find the derivative when (x,y)=(π,0)​
lisov135 [29]
<h2>Answer:</h2>

Shown in the explanation

<h2>Step-by-step explanation:</h2>

Recall that an implicit function is a relation given by the form:

{\displaystyle R(x_{1},\ldots, x_{n})=0}

Where R is a function of two or more variables. In this case, that function is:

y = sin(x+y)

and is implicit because we can define it as:

y-sin(x+y)=0 having two variables.

So, let's take the derivative:

\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right) \\ \\

Applying chain rule:

\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)

But:

\frac{d}{dx}\left(y\right)=y'

Therefore:

y'=\cos \left(x+y\right)\left(1+y'\right)

Isolating y':

\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}

When (x,y)=(\pi,0):

\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}

4 0
3 years ago
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abruzzese [7]

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3 years ago
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