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MaRussiya [10]
3 years ago
8

A roll of quarters is 10$ how many q in a roll write a equation

Mathematics
2 answers:
AleksAgata [21]3 years ago
8 0

Answer:40

Step-by-step explanation:

10 devided by .25  = 40

Charra [1.4K]3 years ago
3 0
ANSWER:
40 quarters

EXPLANATION:
There are 4 quarters in a dollar so there are 40 quarters in 10 dollars
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It would be 11.25 inches because 1800 divided by 160 is 11.25
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Find the coordinates of the midpoint
love history [14]

Answer:

(2,2)

Step-by-step explanation:

(5+-1/2,8+-4/2)

(-2,2)

7 0
3 years ago
You have 4 cards, 2 black and 2 red. You play a game where during each round you draw a card. If it's black, you lose a point. I
sineoko [7]

Answer: The expected value of this game is 2/3

Step-by-step explanation:

Give that

If it's black, you lose a point. If it's red, you gain a point. 

And then you can stop at any time. But you should never stop when you are losing because that can guarantee 0 by drawing all the cards.

Assuming you should stop after three cards when you are +2.

The only question is whether to draw if you are +1 on the first draw.

If you draw red first, You have 1/3 chance of drawing red again and this will give you +2 points

1/3 chance of drawing two blacks and earn zero point, chance of drawing black-red and earn +1. This gives +1, so it doesn't matter whether you draw or not.

From the beginning, If you draw red (probability 1/2 you end +1. If you draw black and then draw two reds (probability 1/6 you end +1) Otherwise you break even with probability 1/3. Overall, the value is 2/3

8 0
3 years ago
Solve for u, z, y, or t.<br>​
jekas [21]

Doubt --!!

Where is your question!??????

5 0
3 years ago
AABC has vertices at A(1, -9), B(8,0), and C(9,-8).
Rainbow [258]

Check the picture below.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

6 0
3 years ago
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