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We are basically given most of what we need to calculate the height of the cannonball.
We use the formula h = –16t+ vt + s to find the height requested.
Let v = 160
Let s = 10
Let t = time in seconds
Was the value of t included? We need to know t to plug into our formula.
You know everything else except for t. Go back to your notes to search for t. Afterward, plug the value of t and everything else given above into the formula and calculate to find h.
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
P(at least 1 missed shot) = 100 - 77.6 = 22.4%
Number of times to expect al least 1 missed shot = 0.224 x 2500 = 560 times.
Answer:
60
Step-by-step explanation:
220*2 = 440
+ 220/2 = 550
