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Lubov Fominskaja [6]

2 years ago

8

A fish was swimming LaTeX: 2\frac{1}{2}2 1 2 feet below the water's surface at 1:00 p.M. Three hours later, the fish was at a de

pth that is LaTeX: 3\frac{1}{4}3 1 4feet below where it was at 1: 00 p.M. What rational number represents the position of the fish with respect to the water's surface at 4:00 p.M.?

1 answer:

marin [14]2 years ago

8 0

Answer:

$\frac{23}{4}$ feet

Step-by-step explanation:

The depth of the fish to the surface of water at 1:00 pm = 2 $\frac{1}{2}$ feet

= $\frac{5}{2}$ feet

The depth of the fish to its initial position at 4:00 pm = 3 $\frac{1}{4}$ feet

= $\frac{13}{4}$ feet

The position of the fish with respect to the water's surface at 4:00 pm = the depth of the fish to the surface of water at 1:00 pm + the depth of the fish to its initial position at 4:00 pm

= $\frac{5}{2}$ + $\frac{13}{4}$

= $\frac{10 + 13}{4}$

= $\frac{23}{4}$

The position of the fish with respect to the water's surface at 4:00 pm is $\frac{23}{4}$ feet.

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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
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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:

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

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

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$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

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:

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

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

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

$\mathbf{Arrangements = 17280}$

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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:

$\mathbf{n =Chemistry + 1}$

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So, the number of arrangements of these books is:

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

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Read more about permutations at:

brainly.com/question/1216161

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