Apoint is chosen at random in the larger circle. what percent of the time will the point be in the shaded region? round to the nearest tenth of a percent.

The dot will land on red 74.8% of the time

Step-by-step explanation:

To start off we need to find the Area of the whole circle. We can calculate this by using the Area of a Circle Formula on the larger circle measurement.

 .... A is Area, and r is radius.

.... rounded to the nearest hundredth

So now we know that the area of the WHOLE circle is . Now we need to find the area of the inner white circle.

.... rounded to the nearest hundredth

Now that we have the area of the whole circle and the area of the smaller inner circle we subtract them to find the area in red.

Lastly, we divide the Area in Red by the total Area to get the percentage of time the dot will land on red.

.... multiply by 100 to turn decimal into a percent.

Finally, we can see that the dot will land on red 74.8% of the time.

I hope this answered your question. If you have any more questions feel free to ask away at .

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!