Alissa is analyzing an exponential growth function that has been reflected across the y-axis. She states that the domain of the
2 answers:
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<span>
it consists of a rectangle, dimensions L = 4ft and W = 3ft = Ar = 4ft x 3ft = 12 ft^2
</span>then, you have one half of the area of the circle that has diameter R = 3ft - radius will be r = 1.5ft the area of that half will be:
Ac = r^2
/ 2
= (1.5 ft)^2 x 3.14 / 2
= 2.25 ft^2 x 3.4/2
= 7.065 ft^2/ 2
= 3.5325ft^2
A = Ar + Ac
A = 12ft^2 + 3.5325ft^2
A = 15.5325ft^2
<span>At the rate of $2 per square foot, the cost will be:
$ 15.5325*2 = 31.07 </span>
<span>
it consists of a rectangle, dimensions L = 4ft and W = 3ft = Ar = 4ft x 3ft = 12 ft^2
</span>then, you have one half of the area of the circle that has diameter R = 3ft - radius will be r = 1.5ft the area of that half will be:
Ac = r^2
= (1.5 ft)^2 x 3.14 / 2
= 2.25 ft^2 x 3.4/2
= 7.065 ft^2/ 2
= 3.5325ft^2
A = Ar + Ac
A = 12ft^2 + 3.5325ft^2
A = 15.5325ft^2
<span>At the rate of $2 per square foot, the cost will be:
$ 15.5325*2 = 31.07 </span>
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.
Answer:
The cosine of 86º is approximately 0.06976.
Step-by-step explanation:
The third degree Taylor polynomial for the cosine function centered at is:
The value of 86º in radians is:
Then, the cosine of 86º is:
The cosine of 86º is approximately 0.06976.