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2 years ago

The variable that assigns a real number value to an event in a sample space is called.

Mathematics

1 answer:

Bas_tet [7]2 years ago

5 0

The variable that assigns a real number value to an event in a sample space is called Random variable.

<h3>What is Random variable?</h3>

A random variable is a variable that can take on many values. This is because there can be several outcomes of a random occurrence. Thus, a random variable should not be confused with an algebraic variable. An algebraic variable represents the value of an unknown quantity in an algebraic equation that can be calculated. On the other hand, a random variable can have a set of values that could be the resulting outcome of a random experiment.

As, A random variable can be defined as a type of variable whose value depends upon the numerical outcomes of a certain random phenomenon. It is also known as a stochastic variable. Random variables are always real numbers as they are required to be measurable.

For example

Suppose 2 dice are rolled and the random variable, X, is used to represent the sum of the numbers. Then, the smallest value of X will be equal to 2 (1 + 1), while the highest value would be 12 (6 + 6). Thus, X could take on any value between 2 to 12 (inclusive). Now if probabilities are attached to each outcome then the probability distribution of X can be determined.

Hence, The variable that assigns a real number value to an event in a sample space is called Random variable.

Learn more about random variable here:

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How many integers are there between 15 and 2a?

Sunny_sXe [5.5K]

Answer:

14-2a

Step-by-step explanation:

Since 15 and 21 are not included: we will have

=15-2a-1

=14-2a

6 0

2 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$