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3 years ago

A Prefeitura da Cidade Feliz doou um

Mathematics

1 answer:

Igoryamba3 years ago

8 0

Aqui temos a seguinte divisao de terreno:

creche + banheiros + academia = 45% + 3% + 12% = 60%

O que sobra: Fazendo a conta, 100 - 60 = 40, restará 40%

No enunciado informa que sobraram 960m².

Logo concluimos que 40% = 960m²

Sendo assim, regra de 3:

m² %

960 -------- 40

X -------- 60

40X = 960 . 60

X = 57600/40

X = 1440

Logo 1440m² é destinado para: creche, banheiros públicos e academia

de ginástica comunitária.

O terreno tem um total de 1440 + 960 = 2400m²

para cada espaço - novamente diversas regra de 3:

→ creche = 45%

m² %

2400 -------- 100

X -------- 45

X = 108000/100 = 1080

→ banheiros públicos = 3%

m² %

2400 -------- 100

X -------- 3

X = 7200/100 = 72

→ academia de ginástica comunitária = 12%

m² %

2400 -------- 100

X -------- 12

X = 28800/100 = 288

provando:

60% = 1440m² (visto acima)

creche - 1080

banheiros - 72

academia - 288

1080 + 72 + 288 = 1440 (60%)

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It explains about the time between events or the distance between two random events is termed the exponential distribution. Here, the occurrence of the events is continuous and also independent. Moreover, the average rate is constant.

The cumulative distribution function of T is obtained below:

From the information given, let the random variable T be the required time to repair a machine follows exponential distribution with parameter λ
with mean. 1/2 hours

That is, E(x) = 1/2 hours.
The parameter of the random variable T is,
E(x) = 1/λ
λ = 1/E(x)
= 1/(1/2)
= 2

The probability density function of T is,
$f(t) = \left \{ {{2e^{-2t} \ \ \ t > 0} \ \atop {0} \ \ \ elsewhere} \ \right.$
The cumulative density function of T is,
FT(t) = P(T <= t)

$= 1 - e^{- \lambda t}$

$= 1 - e^{- 2t}$
The CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise
to obtain the probability that a repair time exceeds

1/2 hours.
(a) The probability that a repair time exceeds 1/2 hours.
From the given information, the CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise

The required probability is,
P(T <= 1/2) = 1 - P(T <= 1/2)
= 1 - [ $= 1 - e^{- 2(1/2)}$ ]
$= e^{-1}$
= 0.3679

om total probability. It can be expected about 36.79% of chance that repair time exceeds

P(X => x) = 1 - P(X < x)
to obtain the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

(b), The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained below:
From the given information, the CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise
The required probability is,
P = P(T => 12.5∩T>12) / P(T>12)
$= e^{- 25 + 24}$

$= e^{- 1}$
= 0.3679
The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained by dividing the
P = P(T => 12.5∩T>12) / P(T>12)
with
P(T>12).