Answer:
D) y=-4x+2
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-2-6)/(1-(-1))
m=-8/(1+1)
m=-8/2
m=-4
y-y1=m(x-x1)
y-6=-4(x-(-1))
y-6=-4(x+1)
y=-4x-4+6
y=-4x+2
Answer:
What do ya need help with?
Step-by-step explanation:
A right triangle can only be formed when you use the 3/4/5 rule.
The 3/4/5 rule is a^2 + b^2 = c^2.
We will use 10 for a, 15 for b, and 20 for c.
a^2 + b^2 = c^2
Plug in 10 for a, 15 for b, and 20 for c.
10^2 + 15^2 = 20^2
10^2 = 100
15^2 = 225
20^2 = 400
100 + 225 = 400
This is not applicable, as 100 + 225 = 325.
Your answer is:
No, 10,15,20 does not form a right triangle.
I hope this helps!
Let x be the number of bags of Nature's Recipe Venison Meal & Rice Canine 20 percent protein, y be the number of bags of Nutro Max Natural Dog Food 27 percent protein and z be the number of bags of PetSmart Premier Oven Baked Lamb 25 percent protein a dog breeder buys. He wants to make 300 pounds of a mix. Then
20x+17.5y+30z=300.
Now
- in 20x pounds of Nature's Recipe Venison Meal & Rice Canine 20 percent protein is 0.2·20x=4x pounds of protein;
- in 17.5y pounds of Nutro Max Natural Dog Food 27 percent protein is 0.27·17.5y=4.725y pounds of protein;
- in 30z pounds of PetSmart Premier Oven Baked Lamb 25 percent protein is 0.25·30z=7.5z pounds of protein;
- in 300 pounds of mix containing 22 percent protein is 0.22·300=66 pounds of protein.
Then 4x+4.725y+7.5z=66.
You get a system

From the first equation
Substitute it into the second equation:

Simplify it:

y must be divided by 20, then
- y=0, x=9, z=10-6=4;
- y=20, x=16, z is not a whole number;
- y=40, x=23, z is not a whole number;
- y=60, x=30, z=10-20-35<0;
- thus, all other possible z will be <0.
Answer: 9 bags with 1st mix, 0 bags with 2nd mix and 4 bags with 3rd mix.
Answer:
a.
.
b. 
Step-by-step explanation:
By the definition, the expected value of a random variable X with probability mass function p is given by
where the sum runs over all the posible values of X. Given a function g, the random variable Y=g(X) is defined. Note that the function g induces a probability mass function P' given by P'(Y=k) = P(X=g^{-1}(k)) when the function g is bijective.
a. Note that for 1/3ln(2)+1/6ln(5) by choosing the function g(x) = ln(x) the expression coincides with E(g(x)), because if Y = g(x) then E(Y) = P'(Y=1)*ln(1)+P'(Y=2)*ln(2)+P'(Y=5)*ln(5) = P(X=1)*ln(1)+P(X=2)*ln(2)+P(X=5)*ln(5).
b. On the same fashion, the function g(x) = xe^{xt} fullfills the expression of E[g(X)]