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
Answer:
<h2>210 cm³</h2>
Step-by-step explanation:
let V represent the volume of the triangular prism
V = base × height
base = (7×6)/2 = 42/2 = 21
height = 10
then
V = 21×10 = 210
Moving left subtracts from the x value the number you move.
In (2,9) the x value is the number 2
You move 1 unit to the left so subtract 1 from 2:
2-1 = 1
You will be at (1,9)
Answer:
2,016.4
Step-by-step explanation: