Anything with a zero denominator is undefined.
You take 56 and divide by 8 (area is the length times width) and get 7
Question:
If a sample of 2 hammer is selected
(a) find the probability that all in the sample are defective.
(b) find the probability that none in the sample are defective.
Answer:
a 
b 
Step-by-step explanation:
Given
--- hammers
--- selection
This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities
Solving (a): Probability that both selection are defective.
For two selections, the probability that all are defective is:




Solving (b): Probability that none are defective.
The probability that a selection is not defective is:

For two selections, the probability that all are not defective is:




The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area =
<span>
<span>
<span>
209.4395102393
</span>
</span>
</span>
segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412
</span>
</span>
</span>
triangle DEF Area =
<span>
<span>
<span>
173.206
</span>
</span>
</span>
segment area = <span>
<span>
209.4395102393
</span>
-173.206
</span>
segment area =
<span>
<span>
<span>
36.2335102393
</span>
</span>
</span>
segment area =
36.23 m
Source:
http://www.1728.org/circsect.htm
Answer:
As x goes to negative infinity, g(x) goes to zero.
As x goes to positive infinity, g(x) goes to zero.
(So the answer is the second option)
Step-by-step explanation:
We have the function 
First, let's look at what happens when we input smaller and smaller numbers

As we can see, as we input smaller and smaller numbers, the answer gets smaller.
Eventually, these fractions will be so small that they will get closer and closer to zero.
This same thing applies to larger and larger numbers, so the end behavior of each side will both be zero.