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Hoochie [10]
3 years ago
13

. Maya drew one highlighter at a time from a container of highlighters. After each draw, the highlighter was replaced. Maya reco

rded the results of 20 draws in the table below. What is the experimental probability of drawing a yellow highlighter?
Mathematics
1 answer:
Tresset [83]3 years ago
5 0
It depends. How many different colors are their? If it's all yellow it's 100%. But if there are other colors like green pink or blue we need to know what colors and how many of each. Does it give any more information?
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Svetlana's hair is 4 centimeters long. Her hair grows 1.5, point, 5 centimeters per month. Svetlana wants her hair to grow so th
Lana71 [14]
He will need 2 more months to reach up 7 centimeters.
4 cm are long and 1.5 cm grows per month.
in 2 moths her hair will grow 3 inches=4+3=7

2x≤7
where x=1.5 cm
5 0
4 years ago
Juan tried 16 free throw attempts playing basketball. He succeeded 9 times what is the experiential probability of not succeedin
Leni [432]

Answer:

The probability of not succeeding next attempt is 7/16

Step-by-step explanation:

Here in this question, we are asked to write the probability of not succeeding in the next attempt given that he attempted 9 success out of 16 trials.

To calculate this probability of not succeeding, we need to get it from the number of unsuccessful throws.

Thus, if 9 throws are successful out of 16, then the number of unsuccessful throws will be 16-9 = 7

So the probability of not succeeding on next attempt will be 7/16

4 0
3 years ago
Use logarithmic differentiation to find dy/dx: y=((x^3)(2x+3)^1/2) / (x-2)^2
sukhopar [10]
y=\frac { { { x }^{ 3 }\left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } }{ { \left( x-2 \right)  }^{ 2 } }

\\ \\ { \left( x-2 \right)  }^{ 2 }\cdot y={ x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 }\cdot y \right)  } =\ln { \left( { x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 } \right)  } +\ln { y } =\ln { \left( { x }^{ 3 } \right)  } +\ln { \left( { \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ 2\ln { \left( x-2 \right)  } +\ln { y } =3\ln { x } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \ln { y } =3\ln { x } -2\ln { \left( x-2 \right)  } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }

\\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)

\\ \\ \frac { dy }{ dx } =\frac { { x }^{ 3 }\sqrt { 2x+3 }  }{ { \left( x-2 \right)  }^{ 2 } } \cdot \left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)
4 0
3 years ago
Read 2 more answers
Please help if you want.
Alexxandr [17]

Answer:

I'm 92% sure its a

8 0
3 years ago
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Jamie has $20 to spend at a carnival on games and rides. Games cost $1. Rides are $1 each. Write an equation to represent the re
UkoKoshka [18]

Answer: x+y=20

Step-by-step explanation:

8 0
3 years ago
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