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
The solution is in the attached file
Answer:
Step-by-step explanation:
a relation that is not a function
<h3><u>The value of the greater number is 15.</u></h3><h3><u>The value of the smaller number is 7.</u></h3>
x = 2y + 1
3x = 5y + 10
Because we have a value for x we can plug this value in to find the value of y.
3(2y + 1) = 5y + 10
Distributive property.
6y + 3 = 5y + 10
Subtract 5y from both sides.
y + 3 = 10
Subtract 3 from both sides.
y = 7
We can plug this value back into the original equation to find the value of x.
x = 2(7) + 1
x = 15
Answer:
I believe the answer is 20.
Step-by-step explanation:
