Answer: A. As x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞.
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
Make an equation system based on the problem
eg. a is the first number and b is the seond number
An equation for "<span>The sum of two numbers is 53" is
</span>⇒ a + b = 53
An equation for "<span>twice the first number minus three times the second number is 26"
</span>⇒ 2a - 3b = 26
<span>
Solve the equations by elimination and subtitution method
Eliminate a to find the value of b
a + b = 53 (multiplied by 2)
2a - 3b = 26
--------------------------------------
2a + 2b = 106
2a - 3b = 26
------------------- - (substract)
5b = 80
b = 80/5
b = 16
Subtitute the value of b to one of the equations
a + b = 53
a + 16 = 53
a = 53 - 16
a = 37
The numbers are 16 and 37</span>
64% on sale means ( 100-64 = 36) 36% are full price
125 * 0.36 = 45 CDs are full price