6/56 = 3/28........................
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:
Marcela can take up to 13 units.
Step-by-step explanation:
In order to find the number of units that Marcela can take for her college classes, we can set up an inequality and solve for the variable. Since each unit costs $105, we can say that 105u ≤ 1365 where u = the number of units. The number of units multiplied by the cost per unit, must be less than or equal to $1,365. In order to solve for 'u', we can use inverse (opposite) operations and get rid of the coefficient by dividing both sides of the inequality by 105. 1365÷105 = 13. So, the number of units that Marcela can take must be less than or equal to 13 units.
<span>X^2= 1/100
x = 1/10
Because 1/10 * 1/10 = 1/100
</span>
Answer:
50 minutes
Step-by-step explanation:
she spent 1 hr and 30 minutes on homework
she spends 2/3 of an hour on the phone = 40 minutes
1 hr 30 min + 40 minutes = 2 hrs and 10 min
so she had 50 minutes to kill before bedtime since she goes to sleep 3 hrs after dinner
2hr 10 min + 50 min = 3 hrs.
