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Furkat [3]
3 years ago
12

What is the value of sin A? A. 9/12 B. 12/9 C. 9/15 D. 12/15

Mathematics
1 answer:
Whitepunk [10]3 years ago
3 0

D

sinA = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}


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You open a drink stand where you sell coffee and hot chocolate. You sell coffee for $1.80 and hot chocolate for $3.15. At the en
Usimov [2.4K]

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

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7 0
3 years ago
A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa
PSYCHO15rus [73]

Answer:

A) \dfrac{11}{23}

B) \dfrac{88}{345}

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

P_1=\dfrac{22}{46}=\dfrac{11}{23}

B) If we pick 2 athletes at random (without replacement),

  • the probability we get one volleyball player is \dfrac{24}{46}=\dfrac{12}{23};
  • the probability we get another basketball player is \dfrac{22}{45} (only 45 athletes left).

Thus, the probability we get one volleyball player and one basketball player is

P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}

5 0
4 years ago
Gerald buys a bag of 7,500 assorted beads online. A random sample of 150 beads contains 17 red beads. Predict the number of red
Vanyuwa [196]
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.
4 0
3 years ago
Y+8=12 solve please
Nikitich [7]
Y=4
the answer is y=4 because you had to subtract 8 from both sides. 

7 0
3 years ago
Read 2 more answers
What is the answer? please help
amid [387]

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>ab = ( \frac{x1 + x1}{2}  \: , \frac{y1 + y2}{2} ) \\  =  (\frac{2 + 6}{2}  \: , \frac{7 + 3}{2} ) \\  = ( \frac{8}{2}  \:,  \frac{10}{2} ) \\  = (4 ,\: 5)</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>

7 0
3 years ago
Read 2 more answers
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