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3 years ago

1. What is the theoretical probability that the family has two dogs or two cats?

Mathematics

1 answer:

gogolik [260]3 years ago

4 0

let dogs be heads. Let cats be tails. A coin has two sides, in which you are flipping two of them. Note that there can be the possible outcomes

h-h, h-t, t-h, t-t.

How this affects the possibility of two dogs & two cats. Note that there are 1/2 a chance of getting those two (with the others being one of each), which means that out of 4 chances, 2 are allowed.

2/4 = 1/2

50% is your answer

Heads represents cats and tails represents dogs. There is two coins because we are checking the probability of two pets. You have to do the experiment to get your set of data, once you get your set of data, you will be able to divide it into the probability for cats or dogs. To change the simulation to generate data for 3 pets, simply add a new coin and category for the new pet.

Hope this helps you out!

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Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

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