Answer with Step-by-step explanation:
So in this sequence,
1st term = 7
2nd term = 10
3rd term = 13
4th term = 16
10th term = 34
In this sequence,
1st term = -1
2nd term = 3
3rd term = 7
4th term = 11
10th term =35
a clothing manufacturer makes coats, shirts and slacks. The time required for The Cutting, sewing, and packaging on each item is
shown in the table. How many of each should be made to use all available labor hours?
2 answers:
9514 1404 393
Answer:
120 coats, 200 shirts, 180 slacks
Step-by-step explanation:
If we let c, s, p represent quantities of coats, shirts, pants, respectively, then the total number of minutes required for cutting, sewing, and packaging can be expressed by ...
20c +15s +10p = 7200
60c +30s +24p = 17520
5c +12s +6p = 4080
These equations can be written in "augmented matrix" form and submitted to your friendly local matrix solver (online or on your calculator). You would ask for "reduced row-echelon form". The calculator function that accomplishes this is often called RREF. The matrix and result are ...
This means production should be of ...
120 coats, 200 shirts, 180 slacks
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Answer:
(a) The probability of overbooking is 0.2135.
(b) The probability that the flight has empty seats is 0.4625.
Step-by-step explanation:
Let the random variable <em>X</em> represent the number of passengers showing up for the flight.
It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.
Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.
Number of reservations = 19
Regular customers = 14
Seats available = 17 - 14 = 3
Remaining reservations, n = 19 - 14 = 5
P (A remaining passenger will arrive), <em>p</em> = 0.52
The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
Thus, the probability of overbooking is 0.2135.
(2)
Compute the probability that the flight has empty seats as follows:
P (The flight has empty seats) = P (Less than 3 shows up for the flight)
Thus, the probability that the flight has empty seats is 0.4625.