Answer:
How many of these passwords contain at least one occurrence of at least one of the five special characters?
Part A
49-14 = 35
<span>35/7 = 5 </span>
Part B
5 miles each day
Part C
<span>49+35 = 84 </span>
<span>84+40 = 124, which is a good estimate and it is reasonable</span>
Answer:
Step-by-step explanation:
1) √2 + 2√2 = ( 1 + 2 ) √2 = 3*√2 = 3√2
4) √8 + √2 = √2*2*2 + √2 = 2√2 + √2 = ( 2 +1 ) √2 = 3√2
7) 2√5 - √5 = (2 - 1)√5 = 1 √5
10 3√5 - 2√5 = (3 - 2) √5 = 1√5
10 is the answer.
You know that there are 4 numbers and you have to add the numbers then divide them by four to get the mean. So all you have to do is reverse the steps, multiply 30 by 4 to get 120, then add 12+50+48 to get 110. Then subtract, 120-110=10. To check if it is correct add 12, 50, 48, and 10 to get 120, then divide by 4 to get 30. And voila, there is the answer. I hope this helped for future problems like this one.
Answer: The required probability is 
Step-by-step explanation:
Since we have given that
Divisors of 64= 1,2,4,8,16,32,64
Sum would be less than 32
Since it has 7 divisors in total.
If we select any two distinct divisors from first five divisors, the sum will always less than 32.
So, the probability that their sum will be less than 32 is given by

Hence, the required probability is 