Answer:
C.
Step-by-step explanation:
5/7=.714285714 and so on
7/9=.777777778 and so on
Answer:
0.016062.
Step-by-step explanation:
Assume that the population is normal:
~(,
2
) = (1185, 702
).
Then the distribution of the sample mean
̅ =
1 + 2 + 3 + ⋯ + 100
100
is exactly normal with mean
̅= (̅) = = 1185 hours
and standard deviation
̅= (̅) =
√
=
70
√100
= 7 hours.
The standardized variable
=
̅ − ̅
̅
=
̅ − 1185
7
Is distributed as (0,1).
The following value of corresponds to the value ̅= 1200 of ̅:
=
̅−̅
̅
=
1200−1185
7
= 2.142857.
Therefore,
(̅ ≥ 1200) = (
̅ − ̅
̅
≥
1200 − ̅
̅
) = ( ≥
1200 − 1185
7
) = ( ≥ 2.142857) =
= 1 − ( < 2.142857) = 1 − 0.983938 = 0.016062,
because using the command
= NORM. S.DIST(2,142857; TRUE)
from Microsoft Excel we can see that
= 2.142857
gives
( < 2.142857) = 0.983938.
Only rarely, just over one time in a hundred tries of 100 light bulbs, would the average life exceed 1200 hours
Answer:
4
Step-by-step explanation:
5/1=5
4-4+4-5+5=4
Answer:
The running time is quadratic (O(n²) )
Step-by-step explanation:
For the set up, we have a constant running time of C. The, a log-linearsorting is called, thus, its execution time, denoted by T(n), is O(n*log(n)). Then, we call n times a linear iteration, with a running time of an+b, for certain constants a and b, thus, the running time of the algorithm is
C + T(n) + n*(a*n+b) = an²+bn + T + C
Since T(n) is O(n*log(n)) and n² is asymptotically bigger than n*log(n), then the running time of the algorith is quadratic, therefore, it is O(n²).
Answer: 22 /5
-3+7y=5x+2y
Step-by-step explanation:
Subtract 2y from each side
-3+7y-2y=5x+2y-2y
-3 +5y = 5x
Add 5 to each side
5y = 5x+3
Divide each side by 5
y = 5x/5 +3/5
y = x +3/5
Let x = -5
y = -5 + 3/5
y = -25/5 +3/5
y = -22/5
Step-by-step explanation:
