Answer:
![[11.5-(2.5*3)]^2=16](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D16)
Step-by-step explanation:
We want to evaluate ![[11.5-(2.5*3)]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2)
Let us evaluate within the parenthesis first:
![[11.5-(2.5*3)]^2=[11.5-(7.5)]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B11.5-%287.5%29%5D%5E2)
![\implies [11.5-(2.5*3)]^2=[11.5-7.5]^2](https://tex.z-dn.net/?f=%5Cimplies%20%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B11.5-7.5%5D%5E2)
We again subtract within the bracket to obtain:
![[11.5-(2.5*3)]^2=[4]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B4%5D%5E2)
This finally gives us:
![[11.5-(2.5*3)]^2=16](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D16)
Either way. The probability of hitting the circle is:
P(C)=Area of circle divided by area of square
P(W)=(area of square minus area of circle divided by area of square
P(C)=(πr^2)/s^2
P(W)=(s^2-πr^2)/s^2
...
Okay with know dimensions, r=1 (because r=d/2 and d=2 so r=1), s=11 we have:
P(inside circle)=π/121 (≈0.0259 or 2.6%)
P(outside circel)=(121-π)/121 (≈0.9744 or 97.4%)
6x+4=x-11
x=-3
needs to be 20 characters
Answer:
a*sqrt(x+b) + c = d
or
a*√(x+b) + c = d
Step-by-step explanation: