Find the variation constant and an equation of variation for the given situation.

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

$\displaystyle y=14\cdot x$

Step-by-step explanation:

<u>Directly Proportion</u>

It's said that y varies directly proportional as x, if:

$y=k\cdot x$

Where k is a constant of proportionality.

We know that y=7 when x=1/2.

Using the above condition, we can find the value of k:

$\displaystyle 7=k\cdot \frac{1}{2}$

Solving for k:

k=14

Thus, the equation is:

$\boxed{\displaystyle y=14\cdot x}$

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Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$