Answer:
The probability that a randomly selected call time will be less than 30 seconds is 0.7443.
Step-by-step explanation:
We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.
Let X = caller times at a customer service center
The probability distribution (pdf) of the exponential distribution is given by;

Here,
= exponential parameter
Now, the mean of the exponential distribution is given by;
Mean =
So,
⇒
SO, X ~ Exp(
)
To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;
; x > 0
Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)
P(X < 30) =
= 1 - 0.2557
= 0.7443
Answer:
The change in temperature per minute for the sample, dT/dt is 71.
°C/min
Step-by-step explanation:
The given parameters of the question are;
The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)
The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min
Given that both dQ/dT and dQ/dt are known, we have;


Therefore, we get;


For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

Answer:
It's (A) and (D) like the person in the comments said. (B) is wrong.
Step-by-step explanation: