If you’re good with geometry can you help me?

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

4 years ago

Question: (look at image above) (giving 25 points, please DONT COMMENT IF U DONT KNOW)

Katen [24]

Answer:

The only two that share both of these qualities are the isosceles triangle and the isosceles right triangle.

8 0

4 years ago

Geometry pls help !!! Find the value of AB.<br> AB = [?]

V125BC [204]

Answer:

AB = 16 Units

Step-by-step explanation:

In the given figure, CD is the diameter and AB is the chord of the circle.

Since, diameter of the circle bisects the chord at right angle.

Therefore, AE = 1/2 AB

Or AB = 2AE...(1)

Let the center of the circle be given by O. Join OA.

OA = OD = 10 (Radii of same circle)

Triangle OAE is right triangle.

Now, by Pythagoras theorem:

$OA^2 = AE^2 + OE^2 \\10^2 = AE^2 + 6^2 \\100= AE^2 + 36\\100-36 = AE^2 \\64= AE^2 \\AE = \sqrt{64}\\AE = 8 \\\because AB = 2AE..[From \: equation\: (1)] \\\therefore AB = 2\times 8\\\huge \purple {\boxed {AB = 16 \: Units}}$

4 0

3 years ago

an angle measures 27.4 more than the measure of its complementary angle what is the measure of each angle

Arisa [49]

Hello!

To solve, we can set up an algebraic expression.

Complementary angles add up to 90°, and in this instance, one angle is 27.4° more than the other. Therefore:

∠1 = x

∠2 = x + 27.4°

∠1 + ∠2 = 90°

x + (x + 27.4°) = 90°

Combine like terms:

2x + 27.4° = 90°

Subtract 27.4 from both sides:

2x = 62.6°

Divide both sides by 2:

x = 31.3°

Therefore, one of the angles is 31.3°. Solve for the measure of the other angle:

(31.3°) + 27.4° = 58.7°

3 0

3 years ago

Rachel babysits the Michaelson’s kids on Saturdays for 4 hours and is paid $12.50 per hour. On Sundays, she babysits at the Smit

Elena L [17]

Answer: Saturday for 4 hrs = $50

Sunday for 4 hours = $58

Step-by-step explanation:

6 0

3 years ago