Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b) 
And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10
Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s
a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t
a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio = 
ratio = 2.8
Pressure is the force applied perpendicular to a surface divided by the area of the surface. Therefore, from the given values, we can easily calculate the area by dividing the force to the pressure. We do as follows:
Area = F / P
Area = 15000 / 190000
Area = 0.0789 m^2
Hope this answers the question. Have a nice day.
Answer:
Explanation:
Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.
Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t
A(0) = 10 g
Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min
The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min
But the concentration is total amount of salt over 350L constant volume
C = A / 350
Therefore our rate of change for salt A' is
A' = 5 - 5A/350 = 5 - A/70
This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is
So 
with A(0) = 10
c + 350 = 10
c = 10 - 350 = -340
<span>B. The properties they have</span>
