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Sati [7]
3 years ago
13

An office receives a shipment of identical boxes containing office supplies. The shipment contains 6 boxes of pencils, 14 boxes

of pens, and 16 boxes of paper. How many additional boxes of paper must be added to the original 36 boxes in order for the probability of randomly selecting a box of paper from this shipment to be exactly 1/2 ?
Mathematics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

4

Step-by-step explanation:

Given that:

There are 6 boxes of pencils, 14 boxes of pens and 16 boxes of paper.

Total number of original boxes = 6 + 14 + 16 = 36

To find:

Number of additional boxes of paper to be added so that the probability of getting a paper box while selecting a random box becomes exactly \frac{1}{2}.

Solution:

First of all, let us have a look at the formula of probability:

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

Here, number of favorable cases will be equal to the number of paper boxes and

Total number of cases will be equal to the total number of boxes available.

Let us first calculate the original probability:

\frac{16}{36}\neq \frac{1}2

If 1 paper box is added, then probability

\frac{17}{37}\neq \frac{1}2

If 2 paper boxes are added, then probability

\frac{18}{38}\neq \frac{1}2

If 3 paper boxes is added, then probability

\frac{19}{39} \neq \frac{1}2

If 4 paper boxes is added, then probability

\frac{20}{40} = \frac{1}2

Therefore, we need to <em>add 4 paper boxes</em>.

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An FBI survey shows that about 30% (i.e., 0.3) of all property crimes go solved. Suppose that in New York City 15 such crimes ar
konstantin123 [22]

Complete question :

An FBI survey shows that about 30% (i.e., 0.3) of all property crimes go solved. Suppose that in New York City 15 such crimes are committed and they are each deemed independent of each other.

a. What is the probability that exactly 3 of these 15 crimes will be solved? b. What is the probability that at most 3 of these 15 crimes will be solved? c. What is the probability that more than 11 of these 15 crimes will be solved?

Answer:

0.17

0.29686

0.000092

Step-by-step explanation:

Given that :

Probability of success (p) = 0.3

Number of cases (n) = 15

1 - p = 1 - 0.3 = 0.7

Usung binomial distribution formula :

a. What is the probability that exactly 3 of these 15 crimes will be solved?

P(x = 3)

Recall:

P(x = x) = nCx * p^x * (1 - p)^(n-x)

P(x = 3) = 15C3 * 0.3^3 * 0.7^12

P(x = 3) = 455 *

P(x = 3) = 0.17

B.) What is the probability that at most 3 of these 15 crimes will be solved?

P( X ≤ 3) = P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

To save computation time, we can using the binomial probability calculator :

P( X ≤ 3) : 0.00474 + 0.03052 + 0.09156 + 0.17004 = 0.29686

c. What is the probability that more than 11 of these 15 crimes will be solved?

P( X > 11) = P(x = 12) + p(x = 13) + p(x = 14) + p(x = 15)

P( X > 11) = 0.000092

7 0
3 years ago
Find the measure of angle ABD if the measure of angle DBC is 42 degrees and the measure of angle ABC is 88 degrees
snow_lady [41]

    First, realize that angle ABC is the sum of the two other angles. Hence, form an equation in which you subtract the measure of one of the angles from angle ABC to get the measure of the other angle -- the one you are solving for.

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A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
Mariana [72]

Answer:

The questions and problem can be chosen in 1260 ways.

Step-by-step explanation:

Given that, a physics exam consists of 6 open-ended problem and  9 multiple choice questions.

The order of choosing does not matter.

So,we use combination to find ways.

The ways to choose  6 multiple choice from 9 is= ^9C_6

                                                                              =\frac{9!}{6!(9-6)!}

                                                                              =84

The ways to 2 open-ended question from 6  is= ^6C_2

                                                                              =\frac{6!}{2!(6-2)!}

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Since pick 6 multiple choice out of 9 and 2 open-ended question out of 6 both are independent we have multiply both to find required ways.

Total number of ways is =(84×15)

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Answer:

See Explanation;

Step-by-step explanation:

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Answer: yes

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