The answer: " x = 68, y = 72 " .
____________________________
Explanation:
________________________________________
46 + (x - 3) + (y - 3) = 180 .
46 + 1(x - 3) + 1(y-3) = 180 .
46 + 1x - 3 + 1y - 3 = 180 .
46 - 3 - 3 + 1x + 1y = 180 .
40 + x + y = 180 ;
Subtract "40" from EACH SIDE of the equation:
______________________________________
40 + x + y - 40 = 180 - 40 ;
to get:
x + y = 140 ;
_____________________________________
Now:
_____________________________________
65 = (x - 3) ;
↔ x - 3 = 65 ;
Add "3" to EACH SIDE of the equation;
x - 3 + 3 = 65 + 3 ;
to get:
x = 68 .
______________________________
Now:
Since: "x + y = 140" ;
Let us plug in our known value, "68" ; for "x" ;
to solve for "y" ;
__________________________________
x + y = 140 ;
68 + y = 140 ;
↔ y + 68 = 140 ;
Subtract "68" from EACH SIDE of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
______________________________________________
y + 68 - 68 = 140 = 68 ;
y = 72 .
______________________________________________
So, solve for "x" and "y".
x = 68, y = 72 .
_______________________________________________
Answer:
Step-by-step explanation:
<u><em>System A and System B are </em></u><u><em>not equivalent</em></u> !!!
Answer:
48
Step-by-step explanation:
x/8+3=9
x/8=9-3
x/8=6
x=6*8
x=48
Answer:
products - -3x and 6x^2
factors - 8 and 2x
Step-by-step explanation:
The corrected parts of the question has been attached to this answer.
Answer:
A) Probability that the error is less than 0.2 mm; P(X < 0.2) = 0.0272
B) Mean Error (E(X)) = 0.6
C) Variance Error (V(X)) = 0.04
D) Answer properly written in attachment (Page 2)
E) P(0<X<0.8) = 0.8192
Step-by-step explanation:
The probability density function of X is;
f(x) = { 12(x^(2) −x^(3) ; 0<x<1
So, due to the integral symbol and for clarity sake, i have attached all the explanations for answers A - D.
E) The probability that the specification for the error to be between 0 to 0.8 mm is met will be;
P(0<X<0.8) = F(0.8) − F(0) =12([(0.8)^(3)] /3] −[(0.8)^(4)]/4]
= 0.8192
So, the probability is 0.8192.
