On Monday, Rachel bought company stock at a price of $39.60 per share. On Tuesday, the share price changed by -$2.35. On Wednesd
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Answer:
I think your functions are ,
and
If yes then then the third function which is .
Step-by-step explanation:
The function where c is a constant has
Domain :
Range : ( 0 , ∞ )
The above range is irrespective of the value of c.
I have attached the graph of each of the function, you can look at it for visualization.
- <em>
⇒ </em>This function is same as
so its range is <em>( 0 , ∞ )</em>.
- <em>
⇒ </em>If we double each value of the function
, which has range ( 0 , ∞ ), but still the value of extremes won't change as 0*2=0 and ∞*2=∞. Therefore the range remains as <em>( 0 , ∞ )</em>.
- <em>
</em> ⇒ If we add 2 to each value of the function
Answer:
Step-by-step explanation:
For the random variable we define the possible values for this variable on this case
. We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:
We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable like this:
For the distribution of we need to take in count that we are finding a conditional distribution.
given
, for this case we see that
, so then exist
ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:
And if we want to find the joint probability we just need to do this:
And if we multiply the probabilities founded we got: