The answer rewritten is z^2/25
Answer:
0
Step-by-step explanation:
given that we roll a fair die repeatedly until we see the number four appear and then we stop.
the number 4 can appear either in I throw, or II throw or .... indefinitely
So X = the no of throws can be from 1 to infinity
This is a discrete distribution countable.
Sample space= {1,2,.....}
b) Prob ( 4 never appears) = Prob (any other number appears in all throws)
= 
where n is the number of throws
As n tends to infinity, this becomes 0 because 5/6 is less than 1.
Hence this probability is approximately 0
Or definitely 4 will appear atleast once.
We have that
case 1) 2x3 + 4x -----------> <span>C. cubic binomial
</span>The degree of the polynomial is 3----> <span>the greater exponent is elevated to 3
</span>the number of terms is 2
<span>
case 2) </span>3x 5 + 3x 4 + x 3--------> <span>A. Quintic trinomial
</span>The degree of the polynomial is 5----> the greater exponent is elevated to 5
the number of terms is 3
<span>
case 3) </span>x 2 + 3----------> <span>B. quadratic binomial
</span>The degree of the polynomial is 2----> the greater exponent is elevated to 2
the number of terms is 2
<span>
case 4) </span>2x 2 + x − 5 A------------> D. quadratic trinomial
The degree of the polynomial is 2----> the greater exponent is elevated to 2
the number of terms is 3
My graphing calculator says g(4) = f(2).
The 4th selection, x=4, is appropriate.
Answer:
hector's toy
Step-by-step explanation:
it traveled .02 m farther