Parker family drove 120 miles on the first day
Step-by-step explanation:
Given:
Total miles driven on the road trip= 180 miles
solution:
Lets assume the distance drove on the first day is x
The total distance covered on the road trip is 180 miles. so,


Hence, x = 120.
The distance travelled on the first day is 120 miles.
Answer: 0.31
Step-by-step explanation:
Let A denotes the event of Tampa Bay Buccaneers will score a touchdown on their opening drive and B denote the event that their defense will have 3 or more sacks in the game.
Given : P(A)=0.14 P(B) = 0.31 P(A or B)=0.14
Formula : P(A and B)= P(A) + P(B) - P(A or B)
Now, the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game will be :-
P(A and B)= 0.14 + 0.31 - 0.14=0.31
Hence, the required probability : 0.31
Answer:

Step-by-step explanation:
The given system is:


Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.
This gives me the system:


We could solve the first equation for
and replace the second
with that.
Let's do that.

Subtract
on both sides:

So we are replacing the second
in the second equation with
which gives us:





Now recall the first equation we arranged so that
was the subject. I'm referring to
.
We can now find
given that
using the equation
.
Let's do that.
with
:



So the solution is (8,-1).
We can check this point by plugging it into both equations.
If both equations render true for that point, then we have verify the solution.
Let's try it.
with
:


is a true equation so the "solution" looks promising still.
with
:


is also true so the solution has been verified since both equations render true for that point.
Answer:
Cardiac output:
Step-by-step explanation:
Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.
To Find : Find the cardiac output.
Solution:
Formula of cardiac output:
---1
A = 3 mg

Do, integration by parts
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B20t%5Cint%7Be%5E%7B-0.6t%7D%20%5C%2Cdt%7D-%5Cint%5B%5Cfrac%7Bd%5B20t%5D%7D%7Bdt%7D%5Cint%20%7Be%5E%7B-0.6t%7D%20%5C%2C%20dt%5Ddt%5D%5E%7B10%7D_0)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-20te%5E%7B-0.6t%7D%7D%7B0.6%7D%2B%5Cfrac%7B20%7D%7B0.6%7D%5Cint%20%7Be%5E%7B-0.6t%7D%20%5C%2Cdt%5D%5E%7B10%7D_0)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-20te%5E%7B-0.6t%7D%7D%7B0.6%7D%2B%5Cfrac%7B20e%5E%7B-0.6t%7D%7D%7B%280.6%29%5E2%7D%5D%5E%7B10%7D_%7B0%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-200e%5E%7B-6%7D%7D%7B0.6%7D%2B%5Cfrac%7B20e%5E%7B-6%7D%7D%7B%280.6%29%5E2%7D%5D%2B%5Cfrac%7B20%7D%7B%280.60%5E2%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5Cfrac%7B20%281-e%5E%7B-6%7D%7D%7B%280.6%29%5E2%7D-%5Cfrac%7B200e%5E%7B-6%7D%7D%7B0.6%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%5Csim%20%7B54.49%7D)
Substitute the value in 1
Cardiac output:
Cardiac output:
Hence Cardiac output: