Answer: (a) 
(b) 
Step-by-step explanation:
(a) P( Bill hitting the target) = 0.7 P( Bill not hitting the target) = 0.3
P( George hitting the target) = 0.4 P(George not hitting the target) = 0.6
Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3
= 0.54
Chances that George hit the target is = 0.4 x 0.3 = 0.12
So given that exactly one shot hit the target, probability that it was George's shot = 
= .
(b) The numerator in the second part would be the same as of (a) part which is 0.12.
The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.
Given that the target is hit,probability that George hit it = 
= =
The value of f(x) when x=6 is A 27
The formula for volume of a cylinder is V=π r^2 h.
So plug in,
320π=<span>π r^2 20
solve (sadmep (opposite of pemdas))
divide pi out
320=r^2 20
divide 20 out
16=r^2
square root of 16 is 4, so the radius is 4.
Since the 2r=d, the diameter is 8.
P.s. lol it thought I was copying an answer from online...
</span>
Answer:
x=14
Step-by-step explanation:
after you take 3x+42 you have to divid by 3 and get
x=14
<span>The equation of a circle with center C=(h,k) and radius r is:
(x-h)^2+(y-k)^2=r^2
In this case the center is the point C=(a,b)=(h,k)→h=a, k=b, then:
(x-a)^2+(y-b)^2=r^2
We can apply the Pythagorean Theorem to find the distance between any point of the circle P=(x,y) and the Center C=(a,b). This distance must be equal to the radius of the circle:
A^2+B^2=C^2, where A and B are the legs of the triangle and C is the hypothenuse.
In this case, according with the figure: The legs of the triangle are:
A=x-a
B=y-b
And the hypothnuse C=r
Then replacing in the Pythagorean Theorem:
(x-a)^2+(y-b)^2=r^2
Equal to the equation of the circle </span>(x-a)^2+(y-b)^2=r^2
