The answer is this:
A. (6,3)
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
1. 2n > 17
n > 7
2. n/5 ≥ 11
n ≥ 55
3. -3y ≤ -18
y ≥ 6
If these are right, please leave a thanks and a rating.
2(3-4a) + 5(a-7) first do the distributive property
6-8a + 5a-35 now combine like terms
-41 - 3a is the final answer
You multiply both by 42: They jog 6 miles per hour
