All categories

3 years ago

4 m 3m 6 m 8 m UE 3 m Your answer

Mathematics

1 answer:

Lelu [443]3 years ago

6 0

I don't understand, could you explain further?

You might be interested in

(Easy) Whoever is fastest gets brainless and extra points.

alekssr [168]

5/15 = 0.33 cm3 …… that is the answer I think of you can answer mine that would be great

6 0

3 years ago

The population in the town of Huntersville is presently 38,300. The town grows at an annual rate of 1.2%. Find the population of

Basile [38]

Hi there! So the situation that we are talking about for this is compounding, because the population is growing.. The formula for compounding something is P(1 + r)^t, with P representing the initial amount, r representing the rate, and t representing the time in years. First off, let's add the rate into 1. 1.2% is 0.012 in decimal form. 1 + 0.012 is 1.012. Now, because we are looking for the population after 4 years. We raise 1.012 to the 4th power. 1.012^4 is 1.04887093274. It's a long decimal, but do not delete this number from your calculator. Now, we multiply that number by 38,300 in order to get 40,171.7567238 or 40,172 when rounded to the nearest whole number. There. The population of Huntersville will be 40,172 people in 4 years.

3 0

3 years ago

Which of these mappings is a function? (PLEASEEE HELP MEEE)))

nirvana33 [79]

Step-by-step explanation:

creo..... que es la z ...

5 0

3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

4 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

4 years ago