The first limit is known as a left-hand limit. We're approaching x = 3 from the left. This means we start with values smaller than x = 3, say x = 2.5 and move closer to x = 3.
Because x is starting off smaller than 3, we use the first piece of the piecewise function. This is the piece that corresponds to x < 3. So we plug x = 3 into that piece to get
2x^2 - x = 2(3)^2 - 3 = 2(9) - 3 = 18 - 3 = 15
So the result for the left-hand limit is 15.
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The right hand limit will have us start on the other side of x = 3. We start at x = 3.5 and move closer to x = 3. So we'll use the

portion.
Plug x = 3 into the second piece to get
3 - x = 3 - 3 = 0
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The result of the left-hand limit was 15
The result of the right-hand limit was 0
The fact that the results do not match up means that the overall limit at x = 3 does NOT exist.
Answer:
The number of weeks required = 216 weeks
Step-by-step explanation:
Given that:
Margin of Error E = 200
Confidence interval = 95% = 0.95
Level of SIgnificance = 1 - C.I
= 1 - 0.95
= 0.05
Standard deviation = 1500
The Critical value for Z :

The number of weeks( i.e the sample size (n) ) required is :



n = 216.09
n ≅ 216 weeks
1. Number of books that were sold after 8 months:
![f(8)=1000 \sqrt[3]{8}+1000 f(8)=1000 * 2 + 100 = 2000+1000=3000](https://tex.z-dn.net/?f=f%288%29%3D1000%20%5Csqrt%5B3%5D%7B8%7D%2B1000%0Af%288%29%3D1000%20%2A%202%20%2B%20100%20%3D%202000%2B1000%3D3000)
Answer:
3000 books2. Graph of the function is in attachment.
The probability of winning the lottery with one ticket is mathematically given as
P(W)=2.99484408*10^{−8}
<h3>
What is the probability of winning the lottery with one ticket?</h3>
Generally, the equation for the probability is mathematically given as
P(1st ball)=1/34
one ball drawn remain
34-1=33
P(2nd ball)=1/33
In conclusion, the probability of winning the lottery with one ticket is
P(W)=1/34*1/33*1/32*1/31*1/30
P(W)=2.99484408*10^{−8}
Read more about probability
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