Step-by-step explanation:
a) 6-9b=3a
First:Subtract
3b=3a
Second: Devide
3b/3a
=(3÷3)(b÷a)
=1b/a
= b/a
B) 6b+2a=3a
First:Transpose (Group the like terms. signs will change.)
-3a+2a=-6b
=-1a=-6b
Second:Divide
=-1a/-6b:(-1÷-6)(a÷b)
=1a/6b
=a/6b
C)a/3=b+2
First: Transpose
a=3+2+b
=a/6+b
Answer:
It honestly does not matter, but I'd choose the ring because I feel like it is more pleasant to look at.
Step-by-step explanation:
We need to find the area of both shapes to compare:
3/4 of a circle:
A = 6^2pi
= 36pi
36pi(0.75) = 27pi
Ring:
A = 36pi
A2 = 3^2pi = 9pi
36pi-9pi = 27pi
As you can see, the two areas are the same.
Hope this helped! :)
Yes.
a test for a function is the vertical line test, so think about drawing a vertical line on top of this picture. if at any point the line is touching the graph more than once, it is not a function. this graph passes the vertical line test
You multiply the number by itself however many times it says for example 2 to the second power 2x2 is 4
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
