What are two numbers multiplied together to get 15 but add to be 12?
1 answer:
So x times y=15
x+y=12
solve
x+y=12
subtract x from both sides
y=12-x
subsitute
xy=15
x(12-x)=15
distribute
12x-x^2=15
subtract (12x-x^2) from both sides
0=x^2-12x+15
use quadratic formula which is
x=
where you have
0=ax^2+bx+c
so
a=1
b=-12
c=15
subsitute
x=
x=
x=
x=
x=6+/-
x=6+
or 6- 
or aprox
x=10.5826 or 1.41742
the numbers are 10.5826 and 1.41742
x+y=12
solve
x+y=12
subtract x from both sides
y=12-x
subsitute
xy=15
x(12-x)=15
distribute
12x-x^2=15
subtract (12x-x^2) from both sides
0=x^2-12x+15
use quadratic formula which is
x=
where you have
0=ax^2+bx+c
so
a=1
b=-12
c=15
subsitute
x=
x=
x=
x=
x=6+/-
x=6+
x=10.5826 or 1.41742
the numbers are 10.5826 and 1.41742
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Answer and Step-by-step explanation:
A) For the model to be a probability distribution, it has to follow two conditions:
1) The probability of each value of the discrete random variable is between, and included, 0 and 1:
2) The sum of all probabilities is 1;
In the table shown, the probabilities are from 0 to 0.3 - between 0 and 1;
Adding the probabilities: 0 + 0.3 + 0.1 + 0.3 + 0.3 = 1
Therefore, this model is a valid probability distribution model.
B) They are discrete because each value correspond to a finite number of possible values.
C) Expected value is calculated by
E(X) = 2.6
D) P(X=3) = P(3), which means probability of 3:
P(X=3) = 0.3
E) P(X<4): probabilities of values of X that are less than 4:
P(X<4) = P(0) + P(1) + P(2) + P(3)
P(X<4) = 0 + 0.3 + 0.1 + 0.3
P(X<4) = 0.7
F) P(X>0) = P(1) + P(2) + P(3) + P(4)
P(X>0) = 0.3 + 0.1 + 0.3 + 0.3
P(X>0) = 1
G) P(X=5) = P(5)
There is no probability of P(5) because the model doesn't "cover" that number.
H) P(X=0) = P(0)
P(X=0) = 0
I) The total area of any density curve is 1 because it represents all the possible values a variable can assume.