6.25/30.5= 0.2049180327868852 = 0.2 Miles per minute
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
b= (Z-m-z)/(x)
Step-by-step explanation:
Z-m=z+bx
Z-m-z=bx [Transpose z of R.H.S to L.H.S] / [Substract z from L.H.S and R.H.S]
(Z-m-z)/(x)=b [Divide by x on both sides i.e, L.H.S and R.H.S]
b= (Z-m-z)/(x)
Hope this helps you.
ANSWER
See attachment
EXPLANATION
The given inequality is
This implies that,
Multiply both sides of the second inequality by -1 and reverse the inequality sign.
The graphical solution to this inequality is shown in the attachment.
241,864,704 many different passwords are possible.
The total characters of the password is six in which it is given that first character should start with W, X, Y or Z. So the first character can be filled in 4 ways.
For next five characters it can be any letter of the alphabet or any numeric digit so it will be 26 + 10 = 36.
So second, third, fourth, fifth and sixth space can be taken by 36 characters.
Therefore the total no. of passwords will be 4 * 36 * 36 * 36 * 36 * 36 that is equal to 241,864,704.
To solve more questions like this
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