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SpyIntel [72]
3 years ago
6

farmer weighed 20 dairy goats in pounds. Of the goats, 10 were males and 10 were females. Males Females First Quartile 29 21 Sec

ond Quartile (Median) 57 48 Third Quartile 78 72 Based on these samples, what generalization can be made?
Mathematics
1 answer:
aliina [53]3 years ago
4 0

Answer:

lkjhgtfr

Step-by-step explanation:

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At midnight the temperature was 30°F by 6 o'clock in the morning it was dropped 15° & by noon increased by 22° what was the
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Answer: 37°F

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Step-by-step explanation:

30-15=15

15+22=37

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Libby needs to cut pieces of oak molding that are 3/4 yard long. She has 2 yards of oak molding. How many pieces can she cut? Ho
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Answer: She can make two 3/4 yards long of oak molding. She will have half of the second one left over.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
What is 55+84=n+76 answer
rodikova [14]
The answer is n = 63.
8 0
2 years ago
Read 2 more answers
Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g
Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also,  $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0
2 years ago
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