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Flauer [41]
3 years ago
7

Rewrite 140 1/8 in radical form

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0
140^1/8 =  \sqrt[8]{140} 
julia-pushkina [17]3 years ago
3 0
a \frac{x}{n}=\sqrt[n]{a^x} 140  \frac{1}{8} =   \sqrt[8]{140^1} 140 \frac{1}{8} =  \sqrt[8]{140} 

                                                        

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I think the answer is letter B
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What is the first digit quotient for 18.6<br><br><br> Please respond ASAP
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Lucy offers to play the following game with Charlie: "Let us show dimes to each other, each choosing either heads or tails. If w
Darya [45]

Answer:

(a)Charlie is right

(b)$0

Step-by-step explanation:

(a)A game is said to be a fair game when the probability of winning is equal to the probability of losing. Mathematically, a game is said to be fair when the expected value is zero.

In the game, the possible outcomes are: HH, HT, TH and TT.

Charlie wins when the outcome is HH, TT

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Lucy wins when the outcome is HT or TH

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Therefore, the game is fair. Charlie is right.

(b)

If the outcome is HH, Lucy pays $3.

If the outcome is HT or TH, Lucy gets $2.

If the outcome is TT, Lucy pays $1.

The probability distribution of Lucy's profit is given below:

\left|\begin{array}{c|c|c|c}$Profit(x)&-\$3&-\$1&\$2\\P(x)&1/4&1/4&2/4\end{array}\right|

Expected Profit

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Lucy's expected profit from the game is $0.

7 0
3 years ago
2^5×8^4/16=2^5×(2^a)4/2^4=2^5×2^b/2^4=2^c<br>A=<br>B=<br>C= <br>Please I'm gonna fail math
aleksley [76]

9514 1404 393

Answer:

  a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)/(a^c) = a^(b-c)

  (a^b)^c = a^(bc)

___

You seem to have ...

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_____

<em>Additional comment</em>

I find it easy to remember the rules of exponents by remembering that <em>an exponent signifies repeated multiplication</em>. It tells you how many times the base is a factor in the product.

  2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}

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Similarly, division cancels factors from numerator and denominator, so decreases the number of times the base is a factor.

  \dfrac{(2\cdot2\cdot2)}{(2\cdot2)}=2\\\\\dfrac{2^3}{2^2}=2^{3-2}=2^1

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