In a class there are 20 students: 3 are 5' 6", 5 are 5'8", 4 are 5'10", 4 are 6', and 4 are 6' 2". A student is chosen at random
2 answers:
Answer:
The student's expected height is 5' 10.1".
Step-by-step explanation:
In a class there are 20 students: 3 are 5' 6", 5 are 5'8", 4 are 5'10", 4 are 6', and 4 are 6' 2".
First we will convert these heights to inches like:
12 inches = 1 feet
5' 6" = = 66
5'8" = = 68 and so on.
So, we get following heights in inches:
66, 68, 70 ,72, 74
Expected height = (3/20)66 + (5/20)68 + (4/20)70 + (4/20)72 + (4/20)74 = 70.1 inches.
Or in feet, it will be : 5' 10.1"
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Answer:
<em>50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.</em>
Step-by-step explanation:
Suppose, the number of Chef's salad is and the number of Caesar salad is
On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.
So, the two constraints are: and
The total number sold has never exceed 100 salads. So, another constraint will be:
According to the graph of the constraints, the vertices of the common shaded region are: and
<em>(Refer to the attached image for the graph)</em>
The lunch stand makes a $.75 profit on each Chef's salad and $1.20 profit on each Caesar salad. So, the profit function will be:
For (40, 35) ,
For (60, 35) ,
For (60, 40) ,
For (50, 50) , <u><em>(Maximum)</em></u>
For (40, 50) ,
Profit will be maximum when and
Thus, 50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.
Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.
Step-by-step explanation:
Since we have given that
Mean rate defects per unit = 0.3
Since we will use "Poisson distribution":
But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.
So,
so, it becomes,
Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.