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3 years ago

the hotel is renting rooms for $137.19per night what would it cost the company to rent two rooms for four nights

Mathematics

2 answers:

kramer3 years ago

8 0

Answer:

1097.52

Step-by-step explanation:

if 137.19 a night then do it times 4 then that for one room the do it times 2 then thats your answer

PilotLPTM [1.2K]3 years ago

3 0

1097.52 because 137.19 x 8

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Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

X to the power of 6 when x=10

Serjik [45]

The answer is 1000000

4 0

4 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$