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.
1/3 is 0.3333333 and you round it by the 3rd three, so it looks like 0.334
Answer:
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
d. z= 1.3322
Step-by-step explanation:
We formulate our hypothesis as
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
According to the given conditions
p`= 31/225= 0.1378
np`= 225 > 5
n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5
p = 0.4 x= 31 and n 225
c. Using the test statistic
z= p`- p / √pq/n
d. Putting the values
z= 0.1378- 0.11/ √0.11*0.89/225
z= 0.1378- 0.11/ √0.0979/225
z= 0.1378- 0.11/ 0.02085
z= 1.3322
at 5% significance level the z- value is ± 1.645 for one tailed test
The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept Ha : p >0.11 and conclude that the data indicates that the 11% of the world's population is left-handed.
The rejection region is attached.
The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.
which appears to be 0.95 and subtracting from 1 gives 0.04998
Answer: 2y=-6.4 = y=−3.2
-4y= -3.2 = y=0.8
Step-by-step explanation:
only answering what i can sorry
