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2 years ago

I need to find the area, I just need to show my work any answers can help.

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

4 0

Area of rectangle: 15 x 7 = 105
area of semicircle: 1/2 (3.14r^2)
radius is half of diameter so it would be 7.5
1/2 (3.14(7.5)^2)
=1/2 (176.71)
=88.355
rounded is 88.4

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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

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3 years ago

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3 years ago

I need to find the area, I just need to show my work any answers can help.​

I need to find the area, I just need to show my work any answers can help.