The number of presale tickets sold is 271
<em><u>Solution:</u></em>
Let "p" be the number of presale tickets sold
Let "g" be the number of tickets sold at gate
<em><u>Given that, total of 800 Pre-sale tickets and tickets at the gate were sold</u></em>
Therefore,
Presale tickets + tickets sold at gate = 800
p + g = 800 ------ eqn 1
<em><u>Given that, number of tickets sold at the gate was thirteen less than twice the number of pre-sale tickets</u></em>
Therefore,
Number of tickets sold at gate = twice the number of pre-sale tickets - 13
g = 2p - 13 ------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
Substitute eqn 2 in eqn 1
p + 2p - 13 = 800
3p -13 = 800
3p = 800 + 13
3p = 813
p = 271
Thus 271 presale tickets were sold
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
The answer is $41.44.
You take $895× 4.63%= $41.4385.
Answer:
Step-by-step explanation:
Answer is in below attachment