Complete question :
The lifetimes of a certain type of calculator battery are normally distributed. The mean lifetime is 400 days, with a standard deviation of 50 days. For a sample of 6000 new batteries, determine how many batteries will last: 360 and 460 days
Answer:
0.67307
Step-by-step explanation:
Given that :
Mean, m = 400
Standard deviation, s = 50
Sample size, n = 6000
Obtain the standardized score :
Zscore =(x - m) / s
For X = 360
P(x < 360)
Zscore =(360 - 400) / 50
Zscore = - 40 / 50
Zscore = - 0.8
P(Z < - 0.8) = 0.21186
For X = 460
P(x < 460)
Zscore =(460 - 400) / 50
Zscore = 60 / 50
Zscore = 1.2
P(Z < 1.2) = 0.88493
P(Z < 1.2) - P(Z < - 0.8)
0.88493 - 0.21186
= 0.67307
Answer:
5>t
Step-by-step explanation:
45 > 12t + 3(t -8) - 6 multiply 3 times t-8
1) 45 > 12t + 3t - 2 - 6 combine alike terms
2) 45 > 15t - 30 isolate the variable by adding 30
3) 75 > 15t divide by 15
4) 5 > t
Answer:
0.6 x 94, 60/100 x 94
Step-by-step explanation:
0.6 = 60% of 1.
60/100 also = 60% of 1
Answer:
it's (6,0) I can't explain though
Questions 5 and 6. You happy
