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emmainna [20.7K]
2 years ago
11

The volume of a can of Pepsi is 0.33liter. There are 12 cans in 1 carton

Mathematics
1 answer:
marusya05 [52]2 years ago
6 0

Answer:

19 liters 800 milliliters

Step-by-step explanation:

0.33 * 12 * 5 = 19.8 liters

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What is the height of the flagpole in the diagram below
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Answer: The height of the flagpole is 12 feet.

Step-by-step explanation:

Let the height of the flagpole be 'H'.

So, from the figure, we can conclude that, the height and shadow of the flagpole is in proportion to the height and shadow of the stick.

Given:

Height of stick is,

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Shadow of the flagpole is,

Now, as these are in proportion, we have

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Hence, the height of the flagpole is 12 feet.

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Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company
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Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

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This means that n = 16

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that p = 0.5

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0

P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002

P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018

P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085

Then

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

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Step-by-step explanation:

.

.

.

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Bnnn t is a great place to

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