Work out the surface area of this cylinder.2 cm4 cm
1 answer:
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ANSWER:
E[X] ≈ m ln m
STEP-BY-STEP EXPLANATION:
Hint: Let X be the number needed. It is useful to represent X by
m
X = ∑ Xi
i=1
where each Xi is a geometric random variable
Solution: Assume that there is a sufficiently large number of coupons such that removing a finite number of them does not change the probability that a coupon of a given type is draw. Let X be the number of coupons picked
m
X = ∑ Xi
i=1
where Xi is the number of coupons picked between drawing the (i − 1)th coupon type and drawing i th coupon type. It should be clear that X1 = 1. Also, for each i:
Xi ∼ geometric P r{Xi = n} =
Such a random variable has expectation:
E [Xi
] = =
Next we use the fact that the expectation of a sum is the sum of the expectation, thus:
m m m m
E[X] = E ∑ Xi = ∑ E Xi = ∑ = m ∑
= mHm
i=1 i=1 i=1 i=1
In the case of large m this takes on the limit:
E[X] ≈ m ln m
Answer:
x = (-1+√33)/4 or x = (-1-√33)/4
Step-by-step explanation:
Use Quadratic formula with a=2, b=1, c=-4
<h3>Look at the picture.</h3>
<em>commutative property</em> a + b = b + a
... (-5 + 1/2 + 5) = ... (-5 + 5 + 1/2)
<em>inverse property of addition</em> a + (-a) = a - a = 0
... (-5 + 5 + 1/2) = ... (0 + 1/2)
<em>identity property of addition</em> a + 0 = 0 + a = a
... (0 + 1/2) = ... 1/2
<em>inverse property of multiplication</em> a/b · b/a = 1
37 · 2 · 1/2 = 37 · 2/1 · 1/2 = 37 · 1
<em>identity property of multiplication</em> a · 1 = 1 · a = a
37 · 1 = 37