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3 years ago

Bob discovers a box with 30 identical, used, rechargeable batteries in his attic.

Mathematics

1 answer:

monitta3 years ago

5 0

Answer:

a) All of them are out of charge = 9.31x10⁻¹⁰

b) 20% of them are out of charge = 5.529x10⁻⁴

Step-by-step explanation:

This problem can be modeled as a binomial distribution since

There are n repeated trials and all of them are independent of each other.

There are only two possibilities: battery is out of charge and battery is not out of charge.

The probability of success does not change with trial to trial.

Since it is given that it is equally likely for the battery to be out of charge or not out of charge so probability of success is 50% or 0.50

P = 0.50

1 - P = 0.50

a) All of them are out of charge?

Probability = nCx * P^x * (1 - P)^n-x

Probability = ₃₀C₃₀(0.50)³⁰(0.50)⁰

Probability = 9.31x10⁻¹⁰

b) 20% of them are out of charge?

0.20*30 = 6 batteries are out of charge

Probability =₃₀C₆(0.50)²⁴(0.50)⁶

Probability = 5.529x10⁻⁴

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3 years ago

A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit

Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = - (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] = 20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

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3 years ago

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skad [1K]

Answer:

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Step-by-step explanation:

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Step 1 of 3

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