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deff fn [24]
2 years ago
10

A bowl of berries contains 2 strawberries, 4 blackberries, 1 raspberry, and 3 blueberries Loue will randomly select two berries

one at a time from the bowl and not put them back. What is the probability that the first berry Louie selects will be a raspberry and the second berry will be a blueberry?
.) 3/9
b) 3/10
c) 1/30
d) 1/90​
Mathematics
1 answer:
Sunny_sXe [5.5K]2 years ago
3 0

Answer

b

Step-by-step explanation:

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(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t&lt;00≤t&lt;11≤t&lt;7t≥7;f(t)={0,t&lt;0−5,0≤t&lt;1−6,1≤t&lt;76,t≥7; 1. wr
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f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
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