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Scorpion4ik [409]
3 years ago
12

Explain how the graphs of the equations below are related to the graph of y = f(x).

Mathematics
1 answer:
Rudiy273 years ago
8 0

Answer: Hello mate!

we have the function y = f(x) and we want to see how the next changes affect the graph of it:

a) y = 5f(x)

this is a "dilatation" in the y-axis, this means that the graph is 5 times higher than the original graph.

b) y = f(x − 4)

Now we have y = f(0) when x = 4, so here we have the whole graph translated to the right by 4 units

c) y = −2f(x)

This case is similar to the case in a, here we have dilatation of 2 units, but also the points where before the graph was positive, now is negative, and where the graph was negative, now is positive.

So we also have a reflection over the x-axis

d)  y = f(3x)

Now, in the point x = 1, we have the value of y = f(3), this means that the value of y = f(1) is in between the x values of 0 and 1, then this is a contraction in the x-axis by 3 times.

e. y = 2f(x) − 5

Ok, here we have two things.

First, you can see another dilatation in the y-axis, but here we also have a subtraction of a constant, this means that the graph is dilated two times in the y-axis, and is translated by -5 units in the y-axis ( translated down by 5 units)

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Ayo runs a fairground game.In each turn, a player rolls a fair dice numbered 1 - 6 and spins a fair spinner numbered 1 - 12.It c
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Using the definition of expected value, it is found that Ayo can be expected to make a profit of £55.8.

The <em>expected value</em> is given by the <u>sum of each outcome multiplied by it's respective probability.</u>

In this problem:

  • The player wins $6, that is, Ayo loses £6, if he rolls a 6 and spins a 1, hence the probability is \frac{1}{6} \times \frac{1}{12} = \frac{1}{72}.
  • The player wins $3, that is, Ayo loses £3, if he rolls a 3 on at least one of the spinner or the dice, hence, considering three cases(both and either the spinner of the dice), the probability is \frac{1}{6} \times \frac{1}{12} + \frac{1}{6} \times \frac{11}{12} + \frac{5}{6} \times \frac{1}{12} = \frac{1 + 11 + 5}{72} = \frac{17}{72}
  • In the other cases, Ayo wins £1.40, with 1 - \frac{18}{72} = \frac{54}{72} probability.

Hence, his expected profit for a single game is:

E(X) = -6\frac{1}{72} - 3\frac{17}{72} + 1.4\frac{54}{72} = \frac{-6 - 3(17) + 54(1.4)}{72} = 0.2583

For 216 games, the expected value is:

E = 216(0.2583) = 55.8

Ayo can be expected to make a profit of £55.8.

To learn more about expected value, you can take a look at brainly.com/question/24855677

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