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2 years ago

What is 2213 divided by 17

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

7 0

Answer:

130.18

Step-by-step explanation:

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If with some kind of a fraction

ira [324]

the answer is 0, or 0/0 because anything times 0 will always be 0 and 0 does not have a half :)))))

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3 years ago

The altitude of a right triangle that meets the hypotenuse at a right angle divides the triangle into smaller triangles that are

SpyIntel [72]

Since the altitude meets the triangle at a right angle it divides it into Similar triangles

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2 years ago

In gfc factor 24-9 and 14x+63

icang [17]

Answer:

7(2x+9)

3(8-3)

Step-by-step explanation:

14x+63

We can factor out a 7

7(2x+9)

24-9

We can factor out a 3

3(8-3)

4 0

3 years ago

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.

6 0

2 years ago

Miles purchased 5/6 lbs. of bananas, Phillip purchased 8/9 lbs. of oranges.

Sav [38]

They purchased a total of 1 13/18 pounds.

5 0

2 years ago

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