Can someone help me with this?

Cosmo’s family has a pool like this.

Verdich [7]

a. The volume of the pool is 27.483 m³

b. The number of litres of water the pool will hold is 27483 L

c. The pool will fill at a rate of 50 L/min in 549.7 min

The pool is a cylinder, so, we need to find its volume

<h3>a. What is volume of the pool?</h3>

Since the pool is a cylinder, its volume is, V = πd²h/4 where

d = width of pool = 5.4 m and
h = height of pool = 120 cm = 1.20 m

So, V = πd²h/4

= π(5.4 m)² × 1.20 m/4

= π × 29.16 m² × 1.20 m/4

= 34.992π m³/4

= 8.748π m³

= 27.483 m³

So, the volume of the pool is 27.483 m³

<h3>b. How many litres of water will the pool hold?</h3>

Since the volume of the pool, V = 27.483 m³ and 1000 Litres = 1 m³

So, the volume of the pool in litres is V = 27.483 m³

= 27.483 × 1 m³

= 27.483 × 1 000 L

= 27483 L

So, the number of litres of water the pool will hold is 27483 L

Since the volume of the pool is V = 27483 L, the time it will take the pool to fill at a rate of 50 L/min is given by t = Volume/rate

= 27483 L/50 L/min

= 549.66 min

≅ 549.7 min

So, how long will it take to fill the pool at a rate of 50 L/min is 549.7 min

Learn more about volume of pool here:

brainly.com/question/11893973

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7 0

2 years ago

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$