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ivolga24 [154]
2 years ago
5

How many leaves on a tree diagram are needed to represent all possible combinations tossing a coin 4 times?

Mathematics
1 answer:
AlekseyPX2 years ago
7 0
1/2 x 1/2 x 1/2 x 1/2 = 1/16

Answer: 16
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A 24 hour digital watch shows 19:29:00 on its face. The first two digitals adicate
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Answer:

271

Step-by-step explanation:

4 hours and 31 minutes = 240 + 31 = 271 minutes

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2 years ago
Determine whether the given statement is true or false.<br> 27 €(14, 16, 18,..., 32)
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Answer:

answer is true because 27 is belong between 14 to 32 so, it true

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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
erastovalidia [21]

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.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

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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