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
Evaluating the expression 11C3 is very simple to do.
If you are assuming that C is the notation for the combination.
11!
11C3= ____
3!(11-3)!
Answer:
$7724
Step-by-step explanation:
Her estimated quarterly income is $73,040/4 = $18,260.
Her estimated tax rate is ...
27% + 12.4% + 2.9% = 42.3%
This tax rate applied to her estimated earnings gives an estimated quarterly tax of ...
$18,260 × 0.423 = $7723.98
≈ $7724 . . . . . taxes are rounded to the nearest dollar
Answer:
No it is not a solution
Step-by-step explanation:
To solve this, you should replace your (x,y) with (8,3)- 8 being your x value and 3 being your y value.
So this inequality, your y value is supposed to be bigger than the value of whatever 3x-4 equals.
So, your inequality should be 3>3(8)-4
3*8=24-4=20
So is 3>(greater than)20? No. Which then means that (8,3) is not a solution because it would make the inequality not true.
Hope this helps!
