Graph f(x) = 2 x + 3 + 2.

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

crimeas [40]

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.

8 0

3 years ago

Carlos mixed 4/15 L of chocolate syrup 1/2 L of milk

mojhsa [17]

Answer:

23/30 L

Step-by-step explanation:

4/15=8/30

1/2=15/30

8+15=23/30

8 0

2 years ago

What is a reasonable estimate for the product 43 times 68

dsp73

43 times 68.

43
x68

8 times 3 is 24. Bring the 2 over the 4.
8 times 4 is 32. Add 4. Equals 36.
Answer so far is 364.

Add a 0 before answering.
3 times 6 is 18. Bring the 1 over the 4.
4 times 6 is 24. Add 4. Equals 28.
Answer so far is 288

Add 364 + 288
Final answer is 652.

7 0

3 years ago

Read 2 more answers

The 6 elephants at the zoo have a combined weight of 11,894 pounds. About how much does each elephant weigh?

Alisiya [41]

Answer:

1982.33 pounds

Step-by-step explanation:

If they all weigh the same amount then you can figure it out by division.

11894/6=1982.33

3 0

3 years ago