Answer:
2.91 x 10¹² sec
Explanation:
d = distance of nearest star, Proxima Centauri = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m
v = speed of new horizon probe = 14 km/hr = 14000 m/s
t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?
Using the equation
d = v t
Inserting the values given
4.3 x 9.46 x 10¹⁵ = (14000) t
t = 2.91 x 10¹² sec
True, I'm not the best when it comes to science, but I'm pretty sure it's this
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:
b.
Explanation: I do not know much about this but the answer that i think it is was b.
The question here would be what is the volume of the room. The density of air that is given has no use. We simply multiply the dimensions given of the room to determine the volume.
<span>43.0m × 18.0m × 15.0m = 11610m^3 ( 3.28 ft / 1 m)^3 = 4.09 x 10^5 ft^3</span>