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N76 [4]
3 years ago
10

Solve this using a number line. 3 1/8 ÷ 1/8

Mathematics
1 answer:
sp2606 [1]3 years ago
4 0
Hey there!!

25 is the answer

Hope this helps. c:
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Write ordinates of following points:<br> ( 3 , 4 ) &amp; ( 5 , -3 )
Ne4ueva [31]

See attachment for math work and answer.

4 0
3 years ago
PLZ HURRY Solve for g.<br><br>g3 – 1 &gt; 2
Phoenix [80]

g3 - 1 > 2

g3 - 1 + 1 > 2 + 1

g3 > 3

\frac{g3}{3} >  \frac{3}{3}

g  > 1

//

If the 3 is an exponent, then the person above me is right.

5 0
3 years ago
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Answer:

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Step-by-step explanation:

I hope it will help u

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3 0
3 years ago
Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
Blizzard [7]

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

5 0
3 years ago
I need someone to do this for me in STANDERD ALGORITHM ONLY ok here's the question: 2.1x82
statuscvo [17]
172.2 it correct because u have use you brain and u will get the answer u have to try first  and put your mind to it when u do that then u will be able to do anything 
5 0
3 years ago
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