The maximum value of the objective function is 26 and the minimum is -10
<h3>How to determine the maximum and the minimum values?</h3>
The objective function is given as:
z=−3x+5y
The constraints are
x+y≥−2
3x−y≤2
x−y≥−4
Start by plotting the constraints on a graph (see attachment)
From the attached graph, the vertices of the feasible region are
(3, 7), (0, -2), (-3, 1)
Substitute these values in the objective function
So, we have
z= −3 * 3 + 5 * 7 = 26
z= −3 * 0 + 5 * -2 = -10
z= −3 * -3 + 5 * 1 =14
Using the above values, we have:
The maximum value of the objective function is 26 and the minimum is -10
Read more about linear programming at:
brainly.com/question/15417573
#SPJ1
Step-by-step explanation:
Solution,
since two of these triangle are similar so,
i) LK/LJ=MN/NO
or,j/8=18/12
or,j=18×8/12
.°. j = 12m
ii)JK/JL=MO/NO
or,10/8=n/12
or,120/8=n
.°.n=15m
Answer:
a) 3.6
b) 1.897
c)0.0273
d) 0.9727
Step-by-step explanation:
Rabies has a rare occurrence and we can assume that events are independent. So, X the count of rabies cases reported in a given week is a Poisson random variable with μ=3.6.
a)
The mean of a Poisson random variable X is μ.
mean=E(X)=μ=3.6.
b)
The standard deviation of a Poisson random variable X is √μ.
standard deviation=S.D(X)=√μ=√3.6=1.897.
c)
The probability for Poisson random variable X can be calculated as
P(X=x)=(e^-μ)(μ^x)/x!
where x=0,1,2,3,...
So,
P(no case of rabies)=P(X=0)=e^-3.6(3.6^0)/0!
P(no case of rabies)=P(X=0)=0.0273.
d)
P(at least one case of rabies)=P(X≥1)=1-P(X<1)=1-P(X=0)
P(at least one case of rabies)=1-0.0273=0.9727
