(x-3)(6x-2)
x*6x + x*(-2) -3*6x (-2)*(-3)
6x^{2} -2x-18x+6
6x^2-20x+6
Answer:
Part 2
A. 10
B. 20
C. 15.4
Part 3
A. 96
B.96
C.24
D. 12
Step-by-step explanation:
I am not sure of the last part but I hope the first to help you
for the area of a triangle it is b*h*1/2 and for a rectangle it is bh
We take the equation <span>d = -16t^2+12t</span> and subtract d from both sides to get
0<span> = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
t = [ -(12) </span><span>± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]</span>
= [- 12 ± √(144-64d) ] / (-32)
= [- 12 ± √16(9-4d)] / (-32)
= [- 12 ± 4√(9-4d)] / (-32)
= 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8
Answer: 1.645
Step-by-step explanation:
Given : The quality control manager of a soda manufacturing company wishes to use a significant level of 0.05 to test whether the variabilities in the amount of soda in the company's 16 OZ bottles is more than the variabilities in the company's 12 OZ bottles.
Let 
represents the variance for 16 OZ bottles and 
represents the variance for 12 OZ bottles.
The Null and Alternative Hypothesis will be :-
, since alternative hypothesis is left-tailed , then the test is left-tailed test.
By using the standard normal distribution table for z, the critical value for this hypothesis:-
