bill is on the school archery team. the target has a center bull's-eye and two rings around the bull's-eye. the table gives the
1 answer:
Given that Bill
is on the school archery team. the target has a center bull's-eye and
two rings around the bull's-eye.
Given the table below that gives the probabilities of outcomes.
The probability <span>that Bill will get the next arrow in the inner or outer ring is given by
P(inner ring) + P(outer ring) = 0.297 + 0.423 = 0.72</span>
Therefore, <span>the probability that Bill will get the next arrow in the inner or outer ring is 0.72</span>
Given the table below that gives the probabilities of outcomes.
The probability <span>that Bill will get the next arrow in the inner or outer ring is given by
P(inner ring) + P(outer ring) = 0.297 + 0.423 = 0.72</span>
Therefore, <span>the probability that Bill will get the next arrow in the inner or outer ring is 0.72</span>
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Given Information:
Area of rectangle = 16 square feet
Required Information:
Least amount of material = ?
Answer:
x = 4 ft and y = 4 ft
Step-by-step explanation:
We know that a rectangle has area = xy and perimeter = 2x + 2y
We want to use least amount of material to design the sandbox which means we want to minimize the perimeter which can be done by taking the derivative of perimeter and then setting it equal to 0.
So we have
xy = 16
y = 16/x
p = 2x + 2y
put the value of y into the equation of perimeter
p = 2x + 2(16/x)
p = 2x + 32/x
Take derivative with respect to x
d/dt (2x + 32/x)
2 - 32/x²
set the derivative equal to zero to minimize the perimeter
2 - 32/x² = 0
32/x² = 2
x² = 32/2
x² = 16
x = ft
put the value of x into equation xy = 16
(4)y = 16
y = 16/4
y = 4 ft
So the dimensions are x = 4 ft and y = 4 ft in order to use least amount of material.
Verification:
xy = 16
4*4 = 16
16 = 16 (satisfied)