Answer:
x=1 y=1
Step-by-step explanation:
Let's use the elimination method.
3x+4y=7
2x-y = 1 (multiply both sides by 4)
8x-4y=4
3x+4y=7
(add both equations together)
11x=11
x=1
Plug x=1 into either of the equations
2(1) - y=1
2-y=1
2=1+y
y=1
(1) For the parabola on the bottom row, the domain would be R and the range would be y ≥ -5
(2) For the hyperbola on the bottom row, the domain would be R\{3} (since there is an asymptote at x = 3) and the range would be R\{4} (since there is an asymptote at y = 4)
(3) For the square root function on the bottom row, the domain would be x ≥ -5 and the range would be (-∞, -2]
(4) For the function to the very right on the bottom row, the domain would be R and the range would be (-∞, -3]
Answer:
Step-by-step explanation:
given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
Clerk I II III total
Pack 0.33 0.23 0.44 1
Defective 0.02 0.025 0.015
Pack&def 0.0066 0.00575 0.0066 0.01895
a) probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate
= P(II clerk) -P(II clerk and defective) = ![0.23-0.00575=0.22425](https://tex.z-dn.net/?f=0.23-0.00575%3D0.22425)
b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)
=0.01895
c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3
=P(clerk 3 and def)/P(defective)
=![\frac{0.0066}{0.01895} \\=0.348285](https://tex.z-dn.net/?f=%5Cfrac%7B0.0066%7D%7B0.01895%7D%20%5C%5C%3D0.348285)
The third box plot is your answer
Answer:
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.