Given:
v = 150d
v represents company's production for the coming year
d represents the number of days
150 is the daily production
The rate of change of the function representing the number of vehicles manufactured for the coming year is CONSTANT (150) , and its graph is a STRAIGHT LINE . So, the function is a LINEAR function.
I hope this helps!
First, 
, and if we multiply 
by 
we get
Subtracting this from the numerator gives a remainder of
Next, 
, and if we multiply by 
we have
and subtracting this from the previous remainder, we end up with a new remainder of
Next, 
, and if we multiply by 
we get
Subtracting from the previous remainder, we get a new remainder of
which contains no more factors of 
, so we're done.
So,
Answer:
x=17/4
Step-by-step explanation:
(x-3)/(x+2)=1/5
cross product
1(x+2)=5(x-3)
x+2=5x-15
5x-x-15=2
4x-15=2
4x=2+15
4x=17
x=17/4
Answer:
(X) 0 1 2 3 4
P(X) 0.17 0.23 0.27 0.24 0.09
F(x) 0.17 0.04 0.65 0.91 1
Step-by-step explanation:
Given that;
(X) 0 1 2 3 4
P(X) 0.17 0.23 0.27 0.24 0.09
cumulative distribution function can be calculated by; be cumulatively up the value of p(x) with the values before it;
so
x F(x)
0 P(X = 0) = 0.17
1 P(X = 0) + P(X = 1) = 0.17 + 0.23 = 0.4
2 P(X = 0) + P(X = 1) + P(X = 2) = 0.17 + 0.23 + 0.27 = 0.65
3 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.17 + 0.23 + 0.27 + 0.24 = 0.91
4 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.17 + 0.23 + 0.27 + 0.24 + 0.09 = 1
Therefore, cumulative distribution function f(x) is;
(X) 0 1 2 3 4
P(X) 0.17 0.23 0.27 0.24 0.09
F(x) 0.17 0.04 0.65 0.91 1
