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3 years ago

Surface Area of a sphere, A grapefruit with C = 14cm.

1 answer:

3 0

Surface area SA: $4\pi r^2$
Circumference C: $2\pi r$ dividing this you get $7=\pi r$ and $r= \frac{7}{\pi}$

Now let's input the radius $r$ into the Surface area formula...
$4\pi r^2\rightarrow 4\pi(\frac{7}{\pi})^2=4\pi(\frac{49}{\pi^2})=\frac{196}{\pi}$

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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

Expand (3x + 1)(x + 2) using the box method of by foiling. Show or explain<br> your steps.

Reika [66]

The result would be 3x^2+7x+2.

If you have any question, tell me and I would be more than happy to help! Thanks.

3 0

3 years ago

In order for Aunt Lucy to save the desired $200,000 at age 65, how much will she need to save each month. Round your answer up t

balandron [24]

Answer:

$520.83

Step-by-step explanation:

7 0

3 years ago

Find the terminal point on the unit circle determined by pi over 4 radians

Pavlova-9 [17]

Pi/4 radians = 45 degrees
The terminal point is
((Sqrt2)/2 , (sqrt2)/2)

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3 years ago

In arithmetic what variables look like

Sati [7]

Variables are used in algebra and other branches of mathematis.

They are represented by letters. Usually ( but not necesserally) the letters at the end of the alphabet like x,y and z.

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3 years ago

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