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2 years ago

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Before a renovation, a movie theater had 111 seats. after the renovation, the theater has 144 seats. what is the approximate per

centage increase of the number of seats in the theater? if necessary, round to the nearest tenth of a percent.

1 answer:

Blababa [14]2 years ago

3 0

Change in number of seats = 144 - 111 = 33

percentage increase = 33/111 x 100 = 29.8% (nearest tenths)

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Which expression represents the lateral surface area of the cone?

Brrunno [24]

Answer:

Option "3" is the correct answer to the following question:

Step-by-step explanation:

Given:

Radius of cone (r) = 6 centimeter

height of cone (h) = 8 centimeter

slant height of cone (l) = 10 centimeter

Find:

Lateral surface area of the cone = ?

Computation:

⇒ Lateral surface area of the cone = $\pi$ rl

⇒ Lateral surface area of the cone = $\pi$ (6 centimeters) (10 centimeters)

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2 years ago

A theatre has the capacity to seat people across two levels, the Circle and

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Answer: $76.19\%$

Step-by-step explanation:

<h3> The complete exercise is: " A theatre has the capacity to seat people across two levels, the Circle, and the stalls. The ratio of the number of seats in the circle to a number of seats in the stalls is 2:5. Last Friday, the audience occupied all the 528 seats in the circle and $\frac{2}{3}$ of the seats in the stalls. What is the percentage of occupancy of the theatre last Friday?"</h3>

Let be "s" the total number of seats in the Stalls.

The problem says that the ratio of the number of seats in the Circle to the number of seats in the Stalls is $2:5$ .

Since the number of seats that were occupied last Friday was 528 seats, we can set up the following proportion:

$\frac{2}{5}=\frac{528}{s}$

Solving for "s", we get:

$s*\frac{2}{5}=528\\\\s=528*\frac{5}{2}\\\\s=1,320$

So the sum of the number of seats in the Circle and the number of seats in the Stalls, is:

$Total=1,320\ seats+528\ seats=1,848\ seats$

We know that $\frac{2}{3}$ of the seats in the Stalls were occupied. Then, the number of seat in the Stalls that were occupied is:

$(1,320)(\frac{2}{3})=880$

Therefore, the total number of seats that were occupied las Friday is:

$Total\ occupied=880\ seats+528\ seats=1,408\ seats$

Knowing this, we can set up the following proportion, where "p" is the the percentage of occupancy of the theatre last Friday:

$\frac{100}{1,848}=\frac{p}{1,408}$

Solving for "p", we get:

$(1,408)(\frac{100}{1,848})=p\\\\p=76.19\%$

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1 year ago

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2 years ago

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Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate

mars1129 [50]

Answer: In the beginning he was given 27 sweets.

Step-by-step explanation: The most logical thing to do is to solve it backwards, that is, from what he had at the end of the third day up till the beginning of the first day.

On the third day he ate one-third and had 8 sweets left over. To determine how many he started with on the third day, let the total on day three be called a. If one-third of a is eaten, then the left over which is two-thirds is 8. That is;

8/a = 2/3

By cross multiplication we now have

8 x 3 = 2a

24/2 = a

a = 12

Let the number of sweets he had on day two be called b. If he ate one-third of b and he had 12 left over, then the two-thirds left over is 12 and we now have;

12/b = 2/3

By cross multiplication we now have

12 x 3 = 2b

36 = 2b

36/2 = b

b = 18

Let the number of sweets he had on day one be called x. If he ate one-third of x and he had 18 left over, then the two-thirds left over is 18, and we now have;

18/x = 2/3

By cross multiplication we now have

18 x 3 = 2x

54 = 2x

x = 27

Therefore Tim was given 27 sweets at the beginning.

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2 years ago

Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim

vivado [14]

The average speed of Joshua during that time is 2500 m/h.

Explanation:

It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.

The total time taken by Joshua from 5:15 pm to 8:09 pm is

$2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}$

Dividing we get,

$2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}$

Adding, we have,

$2 { h } 54 \text { min }=2.9 {h}$

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$speed=\frac{distance}{time}$

where $distance=7250$ and $time=2.9$

Hence, substituting the values we have,

$speed=\frac{7250}{2.9}$

Dividing, we get,

$speed=2500$

Thus, the average speed of Joshua during that time is 2500 m/h.

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2 years ago