Answer:
solve one equation for a vriable then substitute that varibale into the other equation.
Step-by-step explanation:
Answer:
See below ~
Step-by-step explanation:
The length and width are <u>x + 6</u> and <u>x</u> respectively.
Equating to the area :
- Area = length × width
- 112 = (x +6)(x)
- 112 = x² + 6x
- x² + 6x - 112 = 0
- x² + 14x - 8x - 112 = 0
- x(x + 14) - 8(x + 14) = 0
- (x - 8)(x + 14) = 0
- x = 8, because distances are always positive
Dimensions :
- Length = x + 6 = 8 + 6 = <u>14 mm</u>
- Width = x = <u>8 mm</u>
Answer:

And we can calculate this with the complement rule like this:

And using the cdf we got:
![P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480](https://tex.z-dn.net/?f=%20P%28X%3E2%29%20%3D%201-%20%5B1-%20e%5E%7B-%5Clambda%20x%7D%5D%20%3D%20e%5E%7B-%5Clambda%20x%7D%20%3D%20e%5E%7B-%5Cfrac%7B1%7D%7B2.725%7D%20%2A2%7D%3D%200.480)
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:
And 0 for other case. Let X the random variable of interest:

Solution to the problem
We want to calculate this probability:

And we can calculate this with the complement rule like this:

And using the cdf we got:
![P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480](https://tex.z-dn.net/?f=%20P%28X%3E2%29%20%3D%201-%20%5B1-%20e%5E%7B-%5Clambda%20x%7D%5D%20%3D%20e%5E%7B-%5Clambda%20x%7D%20%3D%20e%5E%7B-%5Cfrac%7B1%7D%7B2.725%7D%20%2A2%7D%3D%200.480)
Answer:
The approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 inches = 0.15866
Step-by-step explanation:
The complete question is presented in the attached image to this answer.
It is stated that the distribution of tree diameters is approximately normal, hence, this is a normal distribution problem with
Mean diameter = μ = 8 inches
Standard deviation = σ = 2.5 inches
The approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 inches = P(x < 5.5)
To solve this, we first normalize or standardize 5.5 inches
The standardized score for 45mg/L is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (5.5 - 8)/2.5 = - 1.00
The required probability
P(x < 5.5) = P(z < -1.00)
We'll use data from the normal probability table for these probabilities
P(x < 5.5) = P(z < -1.00) = 0.15866
Hope this Helps!!!
Answer:
(x, y, z) = (-4, 29, 17)
Step-by-step explanation:
These three equations have a unique solution. If you want "z arbitrary", you need to write a system of two equations with three variables (or, equivalently, a set of dependent equations).
It is convenient to let a graphing calculator, scientific calculator, or web site solve these.
_____
You can reduce the system to two equations in y and z by ...
subtracting the last equation from the first:
3y -7z = -32
subtracting twice the last equation from the second:
3y -2z = 53
Subtracting the first of these from the second, you get ...
5z = 85
z = 17
The remaining variable values fall out:
y = (53+2z)/3 = 87/3 = 29
x = -9 +2z -y = -9 +2(17) -29 = -4
These equations have the solution (x, y, z) = (-4, 29, 17).