Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
Fractions less than 1/2 are 1/6 & 4/10.
Fractions greater than 1 are 5/4 & 15/12.
Fractions closer to 0 than to 1 are 1/8, 1/4, & 3/10.
Answer:
x + 2y = -8 = y = -4 - 1/2x
y= -1/2x + 4 = y = 4
y= -1/2x - 4 = y = -4
y=-2x - 4 = y = -4
y=1/2x-4 = y = -4
Step-by-step explanation:
(x + 2y = -8 = y = -4 - 1/2x)
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)]
B.) y is vertical x is horizontal (x,y) x meets at -3 and y never touches <span />
