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Ksenya-84 [330]
3 years ago
14

P3. (3+5 points) Let k be any natural number. (a) Prove that for every positive integer n, we have σk(n) = σ−k(n)n k . Conclude

that n is a perfect number exactly when σ−1(n) = 2. (b) Prove that for all positive integers n, we have σ1(n) ≤ n log(n + 1) + γn, where γ is Euler’s constant defined in class. (In this course, log x = loge x denotes the natural logarithm). g
Mathematics
1 answer:
Mrac [35]3 years ago
8 0

Answer:

what????????????

Step-by-step explanation:

what is this my guy

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1 positive x-tile, 12 negative x-tiles, three, and (x -3) (x + 4) are the answers

Step-by-step explanation:

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Jonathan's preschool class has 9 girls and 6 boys. Which proportion can be used to determine the percent, P, of boys in class?
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you are comparing percent to percent and total boys to total class.

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The National Safe Kids Campaign and Bell Sports sponsored a study that surveyed 8,159 children ages 5 to 14 who were riding bicy
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8 0
3 years ago
Which is the side length of a cube with a volume of 1331 m3?
suter [353]
Let's remember the equation for the volume of a cube:
V= x^3 where x is the length of one side
Since a cube has equal length for length, width, and height.

Now, use what you're given
V= 1331

And put that in terms of x
x^3 = 1331

Now solve for x!
x= cube root (1331)

Think about it! cube root of 1331 * cube root of 1331 * cube root of 1331... will equal 1331 m^3!

Hope this helps
8 0
2 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
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