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3 years ago

20 POINTS PLEASE I NEED THIS NOW!!

Mathematics

1 answer:

Andrej [43]3 years ago

3 0

Answer: The answer would be $50. Hope this helps :)

Step-by-step explanation:

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A box contains 4 black shirts, 8 blue shirts, 4 black pants, and 10 blue pants. Determine the probability of randomly selecting

Rashid [163]

<h3>Answer: 11/13</h3>

============================================================

Explanation:

Add up the four numbers given: 4+8+4+10 = 26 items total

Kick out the 4 black shirts since they are neither blue, nor are they pants.

So we have 26-4 = 22 items that are blue or a pair of pants (or both).

Divide that over the total (26) to get the probability we're after:

22/26 = (11*2)/(13*2) = 11/13

Side note: 11/13 = 0.8462 = 84.62% approximately

7 0

2 years ago

10.1 and 10.5 witch one is greater

OLga [1]

Answer:

10.5

Step-by-step explanation:

Because the digit at tenths place in 10.5 is greater than that of in 10.1

$5 > 1$

3 0

3 years ago

Read 2 more answers

The length of a garden can be expressed as 2X +6 feet and the width is X +7 feet what is the perimeter of the garden what is the

Virty [35]

Answer:

6x + 26 feet

Step-by-step explanation:

Assuming the garden is a rectangular shape, we have four sides, where two sides are 2x + 6 feet and two sides are x + 7 feet. To find the perimeter, we add up the lengths of all four sides.

2x + 6 + 2x + 6 + x + 7 + x + 7

2(2x + 6) + 2(x + 7)

6x + 26

6 0

3 years ago

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:

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

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

$\mathbf{Total = 302400}$