Answer:
A darts player practices throwing a dart at the bull’s eye on a dart board. Her probability of hitting the bull’s eye for each throw is 0.2.
(a) Find the probability that she is successful for the first time on the third throw:
The number F of unsuccessful throws till the first bull’s eye follows a geometric
distribution with probability of success q = 0.2 and probability of failure p = 0.8.
If the first bull’s eye is on the third throw, there must be two failures:
P(F = 2) = p
2
q = (0.8)2
(0.2) = 0.128.
(b) Find the probability that she will have at least three failures before her first
success.
We want the probability of F ≥ 3. This can be found in two ways:
P(F ≥ 3) = P(F = 3) + P(F = 4) + P(F = 5) + P(F = 6) + . . .
= p
3
q + p
4
q + p
5
q + p
6
q + . . . (geometric series with ratio p)
=
p
3
q
1 − p
=
(0.8)3
(0.2)
1 − 0.8
= (0.8)3 = 0.512.
Alternatively,
P(F ≥ 3) = 1 − (P(F = 0) + P(F = 1) + P(F = 2))
= 1 − (q + pq + p
2
q)
= 1 − (0.2)(1 + 0.8 + (0.8)2
)
= 1 − 0.488 = 0.512.
(c) How many throws on average will fail before she hits bull’s eye?
Since p = 0.8 and q = 0.2, the expected number of failures before the first success
is
E[F] = p
q
=
0.8
0.2
= 4.
X = 7.............................
Answer:
PART A: Inequality (a)
Solve for y
The graph of y ≥ ⅓(8-x) is represented by the upper red line and all points in the shaded area below it. The line is solid because points on the line satisfy the conditions.
Inequality (b)
Solve for y
The graph of y ≥ 2 - x is represented by the lower blue line and all points in the shaded area above it. The line is solid because points on the line satisfy the conditions. The solution lies in the purple area. It consists of all combinations of x and y that make y ≥ ⅓(8 - x) and y ≥ 2 - x. A practical but not a mathematical condition is that all values of x and y must be zero or positive numbers (for example, you can't have -2 servings of food), so I have plotted only the numbers in the first quadrant.
PART B: If a point is a solution of the system, then the point must satisfy both inequalities of the system.
For x=8, y=2. Verify inequality A is not true. So the point does not satisfy inequality A. Therefore, the point is not included in the solution area for the system.
PART C: I choose the point (3,1) which is included in the solution area for the system.
That means Michelle buys 3 serves of dry food and 1 serving of wet food.
Step-by-step explanation:
Plz mark Brainliest?
If 
, then by rationalizing the denominator we can rewrite
Now,
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