Answer:
There's two ways to solve this.
Step-by-step explanation:
First way:
Let's divide the width and length by 2.3.
46÷2.3=20
69÷2.3=30
20×30
600 ft²
Second Way:
Let's find the area of the actual room.
46×69
3,174 ft²
Let's find the scale drawing squared.
2.3²=5.29
3,174÷5.29
600 ft²
All you have to do is substitute all the Xs to and get a final y output.
for example:
if we take the number x is -1 all you do is:
y=-4(-1)+2
y=4+2
y=6
thats the first one done
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
Part A:
To determine the values of the times to which the height of the two cannon balls are the same, we equate the given functions.
H(t) = g(t)
Substitute the expressions for each.
-16t² + 48t + 12 = 10 + 15.2t
Transpose all the terms to the left-hand side of the equation.
-16t² + (48 - 15.2)t + (12 - 10) = 0
Simplifying,
-16t² + 32.8t + 2 = 0
The values of t from the equation are 2.11 seconds and -0.059 seconds
Part B:
In the context of the problem, only 2.11 seconds is acceptable. This is because the second value of t which is equal to -0.059 seconds is not possible since there is no negative value for time.
