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3 years ago

Someone help me please

Mathematics

1 answer:

Rudik [331]3 years ago

4 0

BC= 8 cm
The measure of angle C is 90
Cos B = 8/15
Sin C = 1
Tan C = 15/8

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A cooler contains fifteen bottles of sports drink: eight lemon-lime flavored and seven orange flavored

dem82 [27]

Answer:

Mutually exclusive,

$P(\text{Lemon-lime or orange})=\frac{2}{3}$

Step-by-step explanation:

Please consider the complete question:

Determine if the scenario involves mutually exclusive or overlapping events. Then find the probability.

A cooler contains twelve bottles of sports drink: four lemon-lime flavored, four orange flavored, and four fruit-punch flavored. You randomly grab a bottle. It is a lemon-lime or an orange.

Let us find probability of finding one lemon lime drink.

$P(\text{Lemon-lime})=\frac{\text{Number of lemon lime drinks}}{\text{Total drinks}}$

$P(\text{Lemon-lime})=\frac{4}{12}$

$P(\text{Lemon-lime})=\frac{1}{3}$

Let us find probability of finding one orange drink.

$P(\text{Orange})=\frac{\text{Number of orange drinks}}{\text{Total drinks}}$

$P(\text{Orange})=\frac{4}{12}$

$P(\text{Orange})=\frac{1}{3}$

Since probability of choosing a lemon lime doesn't effect probability of choosing orange drink, therefore, both events are mutually exclusive.

We know that probability of two mutually exclusive events is equal to the sum of both probabilities.

$P(\text{Lemon-lime or orange})=P(\text{Lemon-lime})+P(\text{Orange})$

$P(\text{Lemon-lime or orange})=\frac{1}{3}+\frac{1}{3}$

$P(\text{Lemon-lime or orange})=\frac{1+1}{3}$

$P(\text{Lemon-lime or orange})=\frac{2}{3}$

Therefore, the probability of choosing a lemon lime or orange is $\frac{2}{3}$ .

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