Answer:
C
Step-by-step explanation:
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
I do believe it’s 250$
**The first choice:**
Principal: 5,000$
Rate(%): 5
Time(year): 3 years
A: 5,750$
**Second choice**
Principal: 5,000$
Rate(%): 4%
Time(year): 5 years
A: 6,000$
6,000-5,750=250$
Hope this helps!
4n² - 16n - 84 = 0
Multiply both sides by 1/4 :
n² - 4n - 21 = 0
Add 21 to both sides:
n² - 4n = 21
Complete the square by adding 4 to both sides:
n² - 4n + 4 = 25
(n - 2)² = 25
Solve for n :
n - 2 = ± √25
n - 2 = ± 5
n = 2 ± 5
Then n = 2 + 5 = 7 or n = 2 - 5 = -3.
