The minimum amount of points required to make a plane is 3 points.
Answer:
The values of x for which the model is 0 ≤ x ≤ 3
Step-by-step explanation:
The given function for the volume of the shipping box is given as follows;
V = 2·x³ - 19·x² + 39·x
The function will make sense when V ≥ 0, which is given as follows
When V = 0, x = 0
Which gives;
0 = 2·x³ - 19·x² + 39·x
0 = 2·x² - 19·x + 39
0 = x² - 9.5·x + 19.5
From an hint obtained by plotting the function, we have;
0 = (x - 3)·(x - 6.5)
We check for the local maximum as follows;
dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0
6·x² - 38·x + 39 = 0
x² - 19/3·x + 6.5 = 0
x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2
∴ x = 1.288, or 5.045
At x = 1.288, we have;
V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99
V ≈ 22.99 in.³
When x = 5.045, we have;
V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023
Therefore;
V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5
The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.
Answer:
a)0.7
b) 10.03
c) 0.0801
Step-by-step explanation:
Rate of return Probability
9.5 0.1
9.8 0.2
10 0.3
10.2 0.3
10.6 0.1
a.
P(Rate of return is at least 10%)=P(R=10)+P(R=10.2)+P(R=10.6)
P(Rate of return is at least 10%)=0.3+0.3+0.1
P(Rate of return is at least 10%)=0.7
b)
Expected rate of return=E(x)=sum(x*p(x))
Rate of return(x) Probability(p(x)) x*p(x)
9.5 0.1 0.95
9.8 0.2 1.96
10 0.3 3
10.2 0.3 3.06
10.6 0.1 1.06
Expected rate of return=E(x)=sum(x*p(x))
Expected rate of return=0.95+1.96+3+3.06+1.06=10.03
c)
variance of the rate of return=V(x)=![sum(x^2p(x))-[sum(x*p(x))]^2](https://tex.z-dn.net/?f=sum%28x%5E2p%28x%29%29-%5Bsum%28x%2Ap%28x%29%29%5D%5E2)
Rate of return(x) Probability(p(x)) x*p(x) x²*p(x)
9.5 0.1 0.95 9.025
9.8 0.2 1.96 19.208
10 0.3 3 30
10.2 0.3 3.06 31.212
10.6 0.1 1.06 11.236
sum[x²*p(x)]=9.025+19.208+30+31.212+11.236=100.681
variance of the rate of return=V(x)=sum(x²*p(x))-[sum(x*p(x))]²
variance of the rate of return=V(x)=100.681-(10.03)²
variance of the rate of return=V(x)=100.681-100.6009
variance of the rate of return=V(x)=0.0801
Answer:

0 ≤ Ф ≤ 4π.
Step-by-step explanation:
since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))
the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.
We take

Note that
Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.
The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.
3>-2 i believe
djdjjdksksk sorry i needed more characters