Answer:
46mins
EXPLANATION
20 cents=10mins
customers working hrs=$4.60*10
$4.60*10=46 minutes
X ≥ 90 + 3x
You simplify the right side, then subtract 180 from both sides, then multiply by negative one.
Answer:
-3.875
Step-by-step explanation:
Answer:
B. 80 cm^2
Step-by-step explanation:
A = bh/2
A = (FE)(16 cm)/2
A = (10 cm)(16 cm)/2
A = 80 cm^2
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
