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
A = a^2
112 = a^2
sqrt 112 = a
sqrt 16 * sqrt 7 = a
4 sqrt 7 = a
so 4 sqrt 7 is the length of each side
To determine the number of pennies each boy received, you would simply divide the total number of pennies that all 3 boys had which was 84 and divide by the number of boys there, to figure out the number of pennies that each boy should in fact receive.
84 pennies/3 boys = 28 pennies should be given to each boy, if each boy were to have an equal number of pennies.
Answer: 601.2
Step-by-step explanation:
62.4 * 13 = 811.2
811.2 - 210 = 601.2
Answer:
x = - 4
Step-by-step explanation:
-3(x+2)=6+2x+8
-3x - 6 = 14 + 2x (-3 * 2 = -6)
-3x -2x = 14 + 6
-5x = 20
x = 20/ -5
x = -4
