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1 year ago

Found to the nearest tenth 2.26795

Mathematics

1 answer:

Flura [38]1 year ago

4 0

Answer: 2.3

Step-by-step explanation:

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Whats the proper abbreviation for twenty-five micrograms

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3 years ago

Estimate the difference 7 3/4-2/25 Show your work.

crimeas [40]

Answer: 7 3/4 - 2/25 = 767/100 = 7 67/100 = 7.67

Step-by-step explanation:

Conversion a mixed number 7 3/4

to a improper fraction: 7 3/4 = 7 3/4 = 7 · 4 + 3/4= 28 + 3/4 = 31/4

To find a new numerator:

a) Multiply the whole number 7 by the denominator 4. Whole number 7 equally 7 * 4/4 = 28/4

b) Add the answer from previous step 28 to the numerator 3. New numerator is 28 + 3 = 31

c) Write a previous answer (new numerator 31) over the denominator 4.

Seven and three quarters is thirty-one quarters

Subtract: 31/4 - 2/25= 31 · 25/4 · 25 - 2 · 4/25 · 4 = 775/100 - 8/100 = 775 - 8/

100 = 767/100

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2 years ago

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Jennifer is saving money to buy a bike. The bike costs $245. She has $125 saved, and each week she adds $15 to her saving. How l

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15x/15= 120/15

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3 years ago

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