Answer: The required number of passwords that can be created is 175760.
Step-by-step explanation: Given that a company needs temporary passwords for the trial of a new payroll software.
Each password will have one digit followed by three letters and the letters can be repeated.
We are to find the number of passwords that can be created using this format.
For the one digit in the password, we have 10 options, 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Since there are 26 letters in English alphabet and letters can be repeated, so the number of options for 3 letters is
26 × 26 × 26 = 17576.
Therefore, the total number of ways in which passwords can be created using the given format is
Thus, the required number of passwords that can be created is 175760.
Answer:
6.6%
Step-by-step explanation:
30% of 22 is 6.6%
Feel free to give brainliest.
Have an awesome day!
Answer:
1 answer which would be -5+5 which =0
Step-by-step explanation:
First you have to get rid of the mixed numbers, so multiply the whole numbers with the denominators and add the numerators so that way you get:
14/3 times 45/4....then cross simplify if possible and finally you can multiply the numerators and the denominators...
7/1 times 15/2 = 105/2....simplify it and get: 52 1/2
Answer:
0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean 490 mg and variance of 400.
This means that 
What is the probability that a randomly selected pill contains at least 500 mg of minerals?
This is 1 subtracted by the p-value of Z when X = 500. So
has a p-value of 0.6915.
1 - 0.6915 = 0.3085
0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals
