
← multiply both the numerator and the denominator by 10

<span>← simplify
</span>

The answer's

.
Answer:
a) The probability is 1/5
bi) The probability each ball is even will be 2/5
bii) The probability the sum of the numbers on the ball is not more than 10 is 16/25
(that's what I think it is)
Answer:
$2147.85
Step-by-step explanation:
199.5*.5=99.75
199.5+99.75=299.25
299.25*.3=89.775
299.25-89.775=209.475
209.475*.04=8.279
209.475+8.379=217.854
Answer:
The answer is 91
Step-by-step explanation:
~Hoped this helped~
Step-by-step explanation:
Given:
and 
We can solve for f(x) by writing

Let 

Then


We know that f(0) = 0 so we can find the value for k:

Therefore,
