Sample space = {p, r, o, b, a, b, i, l, I, t, y} = 11 possible outcomes
1sr event: drawing an I ( there are 2 I); P(1st I) = 2/11
2nd event drawing also an i: This is a conditional probability, since one I has already been selected the remaining number of I is now 1, but also the sample space from previously 11 outcome has now 10 outcomes (one letter selected and not replaced)
2nd event : P(also one I) = 1/10
P(selecting one I AND another I) is 2/11 x 1/10
P(selecting one I AND another I) =2/110 = 0.018
Answer:
This answer is a function because there are no repeating x values. It is okay if there are repeating y values, the x values are the problem. This one has no repeats though so it is a function!
Step-by-step explanation:
Answer:
sorry ... good luck tho.....
Answer:
B) 10
Step-by-step explanation:
|5-15|
Do what is inside the absolute values signs
5-15
-10
|-10|
Absolute value means always positive so take the positive value of -10
10
We know that
circumference=2*pi*r-------> r=circumference/(2*pi)
circumference=30 in
r=circumference/(2*pi)------> 30/(2*pi)-----> 4.7746 in
the radius of a <span>standard basketball is 4.77 in
</span>
<span>the distance between the ball and the rim is equal to
</span>radius of a basketball rim minus radius of a a standard basketball
radius of a basketball rim =18/2------> 9 in
radius of a a standard basketball=4.77 in<span>
</span>
the distance between the ball and the rim=[9-4.77]------> 4.23 in
the answer is
4.23 in
