Using row 4:
<span>coefficients are: 1, 4, 6, 4, 1 </span>
<span>a^4 + a^3b + a^2b^2 + ab^3 + b^4 </span>
<span>Now adding the coefficients: </span>
<span>1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4 </span>
<span>Substitute a and b: </span>
<span>a = 4x </span>
<span>
b = -3y </span>
<span>1(4x)^4 + 4(4x)^3(-3y) + 6(4x)^2(-3y)^2 + 4(4x)(-3y)^3 + 1(-3y)^4 </span>
<span>Now simplify the above: </span>
<span>256x^4 - 768x^3y + 864x^2y^2 - 432xy^3 + 81y^4 </span>
Hello
Answer:
It has open-ended lines that will not form a polygon
Answer:
(a) The probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.
(b) The probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.
Step-by-step explanation:
Let's denote the events as follows:
<em>A</em> = Fell short of expectations
<em>B</em> = Met expectations
<em>C</em> = Surpassed expectations
<em>N</em> = no response
<u>Given:</u>
P (N) = 0.04
P (A) = 0.26
P (B) = 0.65
(a)
Compute the probability that a randomly selected alumnus would say their experience surpassed expectations as follows:
![P(C) = 1 - [P(A) + P(B) + P(N)]\\= 1 - [0.26 + 0.65 + 0.04]\\= 1 - 0.95\\= 0.05](https://tex.z-dn.net/?f=P%28C%29%20%3D%201%20-%20%5BP%28A%29%20%2B%20P%28B%29%20%2B%20P%28N%29%5D%5C%5C%3D%201%20-%20%5B0.26%20%2B%200.65%20%2B%200.04%5D%5C%5C%3D%201%20-%200.95%5C%5C%3D%200.05)
Thus, the probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.
(b)
The response of all individuals are independent.
Compute the probability that a randomly selected alumnus would say their experience met or surpassed expectations as follows:

Thus, the probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.
15% = 0.15
<span>8 = 0.15x </span>
<span>8/0.15 = x </span>
<span>x = 53.33 or 160/3</span>
Answer:
<h2>
36.48 mm^2</h2>
Step-by-step explanation:
The area formula of a circle is:
A=pi*r^2
a semicircle is half the area of a circle:
A=0.5*pi*r^2
now lets find that area:
A=0.5*pi*r^2
A=0.5*pi*8^2
A=0.5*pi*64
A=<u>32pi </u>mm^2
A= around 32*3.14
A= around 100.48 mm^2
Now, lets find the area of the triangle:
it is half the product of the diameter and radius:
A=0.5*8*(8+8)
A=4*16
A=64 mm^2
Now lets find the approximate area of the shaded region:
Asc-At=Ashaded
Ashaded=100.48-64
Ashaded=36.48 mm^2