Answer:
0.05
Step-by-step explanation:
There are five letters to choose from. Since there is no repetition, you can choose one of 5 for the first letter, one of four for the second letter, etc. The total number of five-letter codes is
5 * 4 * 3 * 2 * 1 = 120
Now we need to find out the number of codes that start with M and end with E.
The first position is occupied by M. There is one single letter that can be placed there. The last position is occupied by E. There is a single letter that goes there. There are 3 letters left.
1 * 3 * 2 * 1 * 1 = 6
There are 6 codes that start with M and end with E
p(code that starts with M and ends with E) = 6/120 = 0.05
Answer:
20 dollars
Step-by-step explanation:
sorry if I'm wrong
Your answer is 7 7x7x7+5=26
Answer:
Step-by-step explanation:
From the information given,
Mean, μ = (10.31 + 17.22 + 26.62 + 22.84)/4 = 19.2475
Standard deviation, σ = √summation(x - mean)/n
Summation(x - mean) = (10.31 - 19.2475)^2+ (17.22 - 19.2475)^2 + (26.62 - 19.2475)^2 + (22.84 - 19.2475)^2 = 151.249475
σ = √(151.249475/4)
σ = 6.15
number of sample, n = 4
The z score for 98% confidence interval is 2.33
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
19.2475 ± 2.33 × 6.15/√4
= 19.2475 ± 2.33 × 3.075
= 19.2475 ± 7.16
The lower end of the confidence interval is 19.2475 - 7.16 = 12.09
The upper end of the confidence interval is 19.2475 + 7.16 = 26.41
Since all the terms are odd, the last four terms would be 115+117+119+121
115 + 117 = 232
232 + 119 = 351
351 + 121 = 472
Answer: 472
I hope this helped :)
