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Mekhanik [1.2K]
3 years ago
15

A community hall is shape of a cuboid. The hall is 40 m long, 15 m wide and 3 m high. The community hall needs re-decorating wit

h new point for the walls and ceiling, and new tiles on the floor. A 10 litre tin of paint covers 25m*25m and costs £10. 1m*1m floor tiles cost £3 each. Work out the total cost of paint and tiles needed to re-decorate the community hall.
Mathematics
1 answer:
Licemer1 [7]3 years ago
3 0
Total cost for tiles and paints is $924.
Step-by-step explanation:
We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.
The paint will be required for 4 walls and ceiling.
Let us find area of walls and ceiling.



Therefore, the area of walls and ceiling is 1410 square meters.
Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,


Therefore, the total painting cost is $564.
Tiles will be required for floor. Let us find the area of floor.


Given: 1m squared floor tiles costs $3. So,

Therefore, the total cost for tiles is $360.
Now let us find combined total cost of tiles and paint.

Therefore, the combined total cost of tiles and paint is $924.
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Answer:

b = -9.

Step-by-step explanation:

The line passes through (4, 3) and (7, 12). First, we need to find the slope: the rise over the run.

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Now that we have the slope, we can say that m = 3. So, we have an equation of y = 3x + b. To find b, we can use M(4, 3) and say that y = 3 and x = 4.

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Hope this helps!

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Answer:

Step-by-step explanation:

Hi there!

Let the smaller number be x and the larger number be 2x+1.

According to the question,

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Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
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Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

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∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

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