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Daniel [21]
3 years ago
9

Round 251 to the nearest 100

Mathematics
2 answers:
Olegator [25]3 years ago
7 0
251 rounded to the nearest 100 is 300 because the tens place is 5 or more round up

dangina [55]3 years ago
5 0
251 rounded to the nearest hundred is 300
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2 1/3 divided by 1/4
zloy xaker [14]
Answer:

28/3

Step-by-step explanation:

2 1/3 divided by 1/4

We’ll start by 2 1/3

Multiply 3 x 2 and then add 1

3 x 2 = 6 + 1 = 7

Then add the denominator back

7/3

Now divide 7/3 by 1/4 like this!

Start by multiplying 4 by 7

4 x 7 = 28

Then multiply 3 by 1

3 x 1 = 3

Final results: 28/3
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What is the probability of obtaining 5 heads from 5 coin flips? Give your answer to 5 decimal places.
katovenus [111]

Answer:

0.03125 = 3.125% probability of obtaining 5 heads from 5 coin flips.

Step-by-step explanation:

For each coin flip, there are only two possible outcomes. Either it is heads, or it is tails. The probabilities for each flip are independent from each other. So we use the binomial probability distribution 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.

In this problem we have that:

For each coin toss, heads and tails are equally as likely, so p = \frac{1}{2} = 0.5}

What is the probability of obtaining 5 heads from 5 coin flips?

This is P(X = 5) when n = 5.

So

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

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

0.03125 = 3.125% probability of obtaining 5 heads from 5 coin flips.

6 0
4 years ago
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