The first option 12a - 20
Answer:
475.
Step-by-step explanation:
We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.
We will use z-score formula and normal distribution table to solve our given problem.

Top 60% means greater than 40%.
Let us find z-score corresponding to normal score to 40% or 0.40.
Using normal distribution table, we got a z-score of
.
Upon substituting our given values in z-score formula, we will get:





Therefore, the minimum score necessary to be in the top 60% of the distribution is 475.
Answer:
-80\112 = -5\7
we divided both numbers by 16
Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
; 
