Bill takes a vitamin B pill every 3 days and a vitamin C pill every 4 days. If he takes both pills on day 1, for how many of the
1 answer:
Answer:
6
Step-by-step explanation:
We are given that
Bill takes a vitamin B pills every 3 days and vitamin C pill every 4 days.
We have to find how many of the next 65 days Bill will take both pills on the same day.
To find the number of days on which Bill take both pills on the same day we will find the lcm of 3 and 4.
Lcm (3,4)=
Multiples of 12 lie between 1 and 65 are
12,24,36,48,60
Number of days on which Bill take both pills on the same day=5
If he takes both pills on day 1
Therefore, total number of days on which Bill take both pills on the same day=5+1=6
Hence, otal number of days on which Bill take both pills on the same day=6
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Answer:
1920
Step-by-step explanation:
Answer:
(1). y = x ~ Exp (1/3).
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
Step-by-step explanation:
Kindly check the attachment to aid in understanding the solution to the question.
So, from the question, we given the following parameters or information or data;
(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.
(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "
(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."
(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.
