Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
M represents miles
35+0.10m=25+0.15m
35-25+0.10m=25-25+0.15m
10+0.10m=0.15m
10+0.10m-0.10m=0.15m-0.10m
10=0.05m
m=200
Check:
35+0.10(200)=25+0.15(200)
35+20=25+30
55=55
The solution: m=200 is true.
After 200 miles driven, prices charged by both companies will be the same.
1/4 hour because since it took him 3/4 hr to milk 3 groups, it is reasonable to divide 3/4 by 3. So in 1/4 hour.
Step-by-step explanation:
this is the right answer..hope it helped you .
Answer:
$49,367.5
Step-by-step explanation:
45,500*0.085 = 3,867.5, 45,500 + 3,867.5 = 49,367.5
