Answer:
8% or 0.08
Step-by-step explanation:
Probability of missing the first pass = 40% = 0.40
Probability of missing the second pass = 20% = 0.20
We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:
Probability of missing 1st pass x Probability of missing 2nd pass
i.e.
Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%
Thus, there is 8% probability that he misses two passes in a row.
There is no outliers though 102 and 9 are the furthest numbers from the rest, but not a significant outlier
<span>3 quarters and 2 pennies.3 quarters= $0.75 2 pennies= $0.02= $0.77</span>
Answer:
4
Step-by-step explanation:
And let the missing number be x
We can solve using Algebra
11x - 10 = 34
11x = 34 + 10
11x = 44
x = 44/11
x = 4
For verification:
4(11) - 10 = 34
44 - 10 = 34
34 = 34
Hence, Proved
64/49
thats what I got but im not sure
