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PolarNik [594]
3 years ago
15

A right rectangular prism is packed with cubes of side length fraction 1 over 5 inch. If the prism is packed with 15 cubes along

the length, 6 cubes along the width, and 5 cubes along the height, what is the volume of the prism? (1 point) Group of answer choices A. Fraction 2 and 3 over 10 cubic inches B. Fraction 3 and 3 over 5 cubic inches C. Fraction 7 and 1 over 5 cubic inches D. Fraction 7 and 1 over 10 cubic inches
Mathematics
1 answer:
Ann [662]3 years ago
7 0

Given:

A right rectangular prism is packed with cubes of side length \dfrac{1}{5} inch.

Prism is packed with 15 cubes along the length, 6 cubes along the width, and 5 cubes along the height.

To find:

The volume of the prism

Solution:

Side length of each cube =\dfrac{1}{5} inch.

15 cubes along the length, so length of prism is

l=15\times \dfrac{1}{5}=3\text{ inches}

6 cubes along the width, so width of prism is

b=6\times \dfrac{1}{5}=\dfrac{6}{5}\text{ inches}

5 cubes along the height, so height of prism is

h=5\times \dfrac{1}{5}=1\text{ inch}

Now,

Volume of rectangular prism is

V=l\times b\times h

V=3\times \dfrac{6}{5}\times 1

V=\dfrac{18}{5}

V=\dfrac{3\times 5+3}{5}

V=3\dfrac{3}{5}\text{ cubic inches}

Therefore, the correct option is B.

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A book is 21 centimeters long, how long is this in meters?
Gnesinka [82]

Answer:

0.21 meters :)

Step-by-step explanation:

4 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}

Read more about permutations at:

brainly.com/question/1216161

8 0
2 years ago
Find three consecutive whose sum 237
Nina [5.8K]

Answer:

78 79 80

Step-by-step explanation:

Let the smallest number = x

Let the next number = x + 1

Let the largest number = x + 2

x + x + 1 + x + 2 = 237      Add together to get the equation

3x + 3 = 237                     Subtract 3 from both sides.

3x + 3 - 3 = 237 - 3          Combine the right

3x = 234                           Divide by 3

x = 234/3

x = 78

The smallest number is 78

The next number 79

The highest number is 80

8 0
3 years ago
Read 2 more answers
Which equation does this story match: The temperature is -7. Since midnight the temperature tripled and then rose 5 degrees. Wha
olya-2409 [2.1K]
5x-7=3 :))))))))))))))))))))))
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3 years ago
A recipe calls for 1/2 cup of celery,2 1/3 of water and 3/4 cups of carrots. What the combined amont of these three ingredients?
lys-0071 [83]

Answer:

43/12

Step-by-step explanation:

First we must find a common denominator for all 3 fractions, 1/2 cup, 2 1/3 cup, and 3/4 cup all share 12 as a common denominator.

However, let's convert 2 1/3 cup into an improper fraction: there are 2 (3/3 cups) here multiply (3/3) and (2/1) you get 6/3. Finally, add the remaining 1/3 to 6/3 and you get 7/3.

Now, to get all these fractions to have the same common denominator.

(1/2) ×(6/6)=(6/12)

(7/3)×(4/4)=(28/12)

(3/4)×(3/3)=(9/12)

Add them all together and your answer is: (43/12)

7 0
4 years ago
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