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3 years ago

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There are 3 green balls, 5 red balls, and 2 blue balls in a box. One ball is randomly drawn, replaced and then another ball is d

rawn. What is the probability of getting a green ball then a blue ball? a. 1/15 b. ½ c. 3/100 d. 3/50 (show your work please)

2 answers:

irina [24]3 years ago

5 0

The answer is d. 3/50.

Andre45 [30]3 years ago

4 0

Answer:

Option d. 3/50 is the correct option.

Step-by-step explanation:

Total number of balls given in a box is 10. Out of which different balls are

Green = 3, Red balls = 5, Blue = 2.

One ball was randomly drawn so probability of getting a green ball will be = 3/10

Now the ball was replaced so total number of balls remains the same

Then the probability of getting a blue will be = 2/10

Therefore the probability of both the events simultaneously = (3/10)×(2/10)=6/100 = 3/50

Option d. 3/50 is the answer.

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The legs of the isosceles triangle each measure 14 inches. calculate the length of the hypotenuse

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I'm guessing you are talking about an isosceles RIGHT triangle.

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2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

2 years ago