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pychu [463]
2 years ago
9

For some particular website, the password must start with one of the letters W, X, Y, or Z. Then there must be 5 more charaters,

and each of those characters may be a numeric digit or a letter of the alphabet. How many different passwords are possible?
Mathematics
1 answer:
k0ka [10]2 years ago
3 0

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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Answer:

The probability mass function for the items sold is

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Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

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The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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