Answer:
c = 69 and h = 51
Step-by-step explanation:
You will need two different equations here
use variable c as coffee and variable h as hot chocolate
c + h = 120
1.80c + 3.15h = 284.40
By moving variables, you can alter the first equation
h = 120 - c
Now insert this into the second equation
1.80c + 3.15(120 - c) = 284.40
Multiply
1.80c + 378 - 3.15c = 284.40
Subtract the c
-1.35c + 378 = 284.40
Subtract 378 from both sides
-1.35c = -93.6
Divide both sides by -1.35
c = 69
Then plug into the first equation
69 + h = 120
Subtract 69
h = 51
Answer:
A) 
B) 
Step-by-step explanation:
A survey of 46 college athletes found that
- 24 played volleyball,
- 22 played basketball.
A) If we pick one athlete survey participant at random, the probability they play basketball is

B) If we pick 2 athletes at random (without replacement),
- the probability we get one volleyball player is

- the probability we get another basketball player is
(only 45 athletes left).
Thus, the probability we get one volleyball player and one basketball player is

850. If there is 17 red beads for every 150 beads Then you would divide 7500 by 150 and you get 50. Then multiply 50 with 17 and your answer is 850 red beads. 7500/150=50. 50 x 17= 850.
Y=4
the answer is y=4 because you had to subtract 8 from both sides.
Answer:
<h2>
<em>(</em><em>4</em><em>,</em><em>5</em><em>)</em></h2>
<em>sol</em><em>ution</em><em>,</em>
<em>A</em><em>(</em><em>2</em><em>,</em><em>7</em><em>)</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>></em><em> </em><em>(</em><em>X1,</em><em>y1</em><em>)</em>
<em>B</em><em>(</em><em>6</em><em>,</em><em>3</em><em>)</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>></em><em>(</em><em>x2</em><em>,</em><em>y2</em><em>)</em>
<em>now</em><em>,</em>
<em>
</em>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>