“Determine if Parallel, Perpendicular, or neither , idk if I’m right but I do need help here
1 answer:
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Answer:
Therefore the solution is = k{-7,5,1} where k ∈R
Step-by-step explanation:
Given that,
f₁(x) =x
f₂(x)= x²
f₃(x)= 7x - 5x²
Also,
g(x) = c₁f₁(x)+c₂f₂(x)+c₃f₃(x)
Putting the values of f₁(x), f₂(x) and f₃(x).
g(x) = c₁.x+c₂x²+c₃(7x-5x²)
Given condition that g(x)= 0
∴ c₁.x+c₂x²+c₃(7x-5x²)=0
⇒(c₁+7c₃)x +(c₂-5c₃)x² = 0
Comparing the coefficients of x and x²
∴c₁+7c₃=0 and c₂-5c₃ =0
Let c₃= k [k∈R]
Then c₁ = -7k and c₂=5k
Therefore the solution is = { c₁,c₂,c₃}
= {-7k, 5k, k}
=k{-7,5,1}
Answer:
Step-by-step explanation:
Hello!
a)
The variable of interest is
X: monetary value of a grab bag. ($)
The possible values this variable can take are:
X: 15; 40; 60; 120
To calculate the relative frequency of each possible value of the variable you have to divide the number of times this value is observed by the total of observations (sample size)
There where a total of 300 bags created, from these, 40 cost $40, 20 cost $60, 1 cost $120, and the rest have a monetary value of $15.
There are then 300- (40+20+1)= 239 bags with a monetary value of $15
The frequencies for each value of X are:
X= 15 ⇒ p(X)= 239/300= 0.797
X= 40 ⇒ p(X)= 40/300= 0.133
X= 60 ⇒ p(X)= 20/300=0.067
X= 120⇒ p(X)= 1/300= 0.003
The probability distribution for X is:
Xi: 15; 40; 60; 120
p(Xi):0.797; 0.133; 0.067; 0.003
∑p(Xi)= 0.797+0.133+0.067+0.003=1
b)
To calculate the mean you have to use the following formula:
E(X)= ∑Xp(X)= 15*0.797+40*0.133+60*0.067+120*0.003= 21.655≅ $21.7
c)
The standard deviation is the square root of the variance:
Variance:
V(X)= ∑X²p(X)-(∑Xp(X))²= 676.525 - (21.655)²= 207.585
∑X²p(X)= 15²*0.797+40²*0.133+60²*0.067+120²*0.003= 676.525
Standard deviation:
√V(X)= √207.585= $14.407= $14.41
d)
Considered the probability distribution, it is more likely to buy a $15 worth bag than any of the others, it is therefore not worthwhile to spend $20 in a bag full of merchandise worth $15.
I hope this helps!
