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postnew [5]
3 years ago
11

Y = (x) =(1/8)^x Find f(x) when X = (1/3) Round your answer to the nearest thousandth.

Mathematics
1 answer:
vlada-n [284]3 years ago
6 0

Answer:

1/2.

Step-by-step explanation:

f(x) = (1/8)^x

when x = 1/3

f(x)  =  (1/8)^1/3

f(x) = ∛(1/8)

f(x) = 1 / ∛8

f(x) = 1/2.

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(2x+30y)=(12x+17)<br> Find x
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2x+30y=12x+17
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On December 31, 2016, Osborn Company purchased 30% of Shea Company’s common stock for $220,000. During 2017, Shea Company had a
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the balance of Osborn’s Equity Investment (Shea) account be at the end of 2017 is $233,500

Step-by-step explanation:

Given data

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cash dividends = $30,000

to find out

the balance of Osborn’s Equity Investment (Shea) account be at the end of 2017

solution

we will find out balance of investment i.e. given by formula

balance of investment = Acquistion price + share of income - share of dividend   .................1

so here

share of income = 30% of net income

share of income =30% × 75,000 = $22500    ..............2

and

share of dividend = 30% of cash dividends

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put equation 2 and 3 in equation 1 and we get

balance of investment = Acquistion price + share of income - share of dividend

balance of investment = 220000 + 22500 - 9000

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3 years ago
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In the attached file

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A box contains 5 tickets numbered 1,2,3,4, and 5. Two tickets are drawn at random from the box. Find the chance that the numbers
adelina 88 [10]

Answer:

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) <u>With Replacement</u>

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs

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Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) <u>Without Replacement:</u>

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs

                               =\frac{1}{5}\times \frac{1}{4}\times12\\ =0.60

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

7 0
3 years ago
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