It is 100 degrease I think i am right
This question is incomplete, the complete question is;
Lets consider a hypothetical airline, Mudlark Airlines. On one of its flight, previous records show that about 15% of people who had tickets for the flight did not take the flight. There are 120 seats on the plane.
Suppose that Mudlark decides to sell 140 tickets for this flight. Assuming that all 140 tickets are sold and that the outcome for each ticketholder is independent, how many ticketholders should the airline expect to show up
Answer: the airline should expect 119 ticketholders to show up
Step-by-step explanation:
Given that;
the plane contains 120 seats
Their records shows that 15% (0.15) percent of people who previously buy their flight tickets don't show up.
they sold 140 tickets,
Number of ticketholder expected to show up = ?
Probability of showing up for flight will be;
⇒ 1 - 0.15 = 0.85
Now Expected number of ticketholders to shup will be;
⇒ 140 × 0.85
= 119
Therefore the airline should expect 119 ticketholders to show up
Answer:
The probability of assembling the product between 7 to 9 minutes is 0.50.
Step-by-step explanation:
Let <em>X</em> = assembling time for a product.
Since the random variable is defined for time interval the variable <em>X</em> is continuously distributed.
It is provided that the random variable <em>X</em> is Uniformly distributed with parameters <em>a</em> = 6 minutes and <em>b</em> = 10 minutes.
The probability density function of a continuous Uniform distribution is:
Compute the probability of assembling the product between 7 to 9 minutes as follows:
![=\frac{1}{4}\times [x]^{9}_{7}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Bx%5D%5E%7B9%7D_%7B7%7D%5C%5C)
Thus, the probability of assembling the product between 7 to 9 minutes is 0.50.
Answer:
B
Step-by-step explanation:
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