So u already know that the radius is half of the circle which in this case is 28 mm. And both circles at the top touch both sides of the rectangle.
Process:
28x4=112.
This is the top side, but because you could fit 2 circles at the top and you know the measurement, it is common sense to think that it would be the same on the other side. So the area of the square/ rectangle:
112x112= 12,534 mm
a.
has an average value on [5, 11] of
![f_{\rm ave}=\displaystyle\frac1{11-5}\int_5^{11}(x-7)^2\,\mathrm dx=\frac{(x-7)^3}{18}\bigg|_5^{11}=\frac{4^3-(-2)^3}{18}=4](https://tex.z-dn.net/?f=f_%7B%5Crm%20ave%7D%3D%5Cdisplaystyle%5Cfrac1%7B11-5%7D%5Cint_5%5E%7B11%7D%28x-7%29%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B%28x-7%29%5E3%7D%7B18%7D%5Cbigg%7C_5%5E%7B11%7D%3D%5Cfrac%7B4%5E3-%28-2%29%5E3%7D%7B18%7D%3D4)
b. The mean value theorem guarantees the existence of
such that
. This happens for
![(c-7)^2=4\implies c-7=\pm2\implies c=9\text{ or }c=5](https://tex.z-dn.net/?f=%28c-7%29%5E2%3D4%5Cimplies%20c-7%3D%5Cpm2%5Cimplies%20c%3D9%5Ctext%7B%20or%20%7Dc%3D5)
15x1000+4x100+9x1
I belive that is edpanded notation.
I hope this helps you
regular hexagon =6 equilateral triangle
Area=6.(10.20/2)
Area=600
Answer:
{A and B are independent events}, P(A|B)=P(A)=0.16
Step-by-step explanation:
First of all we need to know when does two events become independent:
For the two events to be independent, P(A|B)=P(A) that is if condition on one does not effect the probability of other event.
Here, in our case the only option that satisfies the condition for the events to be independent is P(A|B)=P(A)=0.16.. Rest are not in accordance with the definition of independent events.