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shepuryov [24]
3 years ago
9

At a carnival, you spin a spinner that has four equal-sized sections, each a different color: green, yellow, red, and blue. If y

ou land on green, you win 2 points. If you land on yellow, red, or blue, you lose 1 point. Write the expected value equation.
What is
a=
b=
c=
d=
e=
f=
g=

Mathematics
2 answers:
Fittoniya [83]3 years ago
7 0

Answer:

Step-by-step explanation:

a=4

b=4

c=-1

d=4

e=-1

f=4

g=-1

dangina [55]3 years ago
6 0

Answer:

second one is

The expected value of the number of points won on a spin is .  -0.25 points.

third one is.

Suppose you spin the carnival spinner 1,000 times. How many points would you expect to win or lose? is lose 250 points

Step-by-step explanation:

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Step-by-step explanation:

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Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)
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Answer:

Given that we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

From trigonometric identities, we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}

1 -  sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}

1 +  sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

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