Answer:
(2,-7)
Step-by-step explanation:
Use the midpoint formula! (Attached is a picture of it!)
1) Plug in the values:
(5-1)/2,(-6+-8)/2
2) Simplify:
4/2,-14/2
3) Solve:
(2,-7)
<u>Answer</u>:
b = 4a
<u>Step-by-step explanation</u>:
<em>changing sides changes sign division to multiplication.</em>
Answer:
Step-by-step explanation:
First confirm that x = 1 is one of the zeros.
f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26
f(1) = 2 - 14 + 38 - 26
f(1) = -12 + 38 = + 26
f(1) = 26 - 26
f(1) = 0
=========================
next perform a long division
x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26
2x^3 - 2x^2
===========
-12x^2 + 28x
-12x^2 +12x
==========
26x -26
26x - 26
========
0
Now you can factor 2x^2 - 12x + 26
2(x^2 - 6x + 13)
The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16
So you are going to get a complex result.
x = -(-6) +/- sqrt(-16)
=============
2
x = 3 +/- 2i
f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)
The zeros are
1
3 +/- 2i
Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
