Division<span>. If two </span>powers<span> have the </span>same base<span> then we can </span>divide<span> the </span>powers<span>. When we </span>divide powers<span> we </span>subtract<span> their </span>exponents<span>. A negative </span>exponent<span> is the </span>same<span>as the reciprocal of the positive </span><span>exponent</span>
Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)
a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.
A(0,0,0)
Reflecting
A(0,0,0)
B(0,5,0)
Reflecting
B(0,-5,0)
C(3,5,0)
Reflecting
C(3,-5,0)
D(3,0,0)
Reflecting
D(3,0,0)
b.)
A(0,0,0)
Moving
A(-2,-3,1)
B(0,-5,0)
Moving
B(-2,-8,1)
C(3,-5,0)
Moving
C(1,-8,1)
D(3,0,0)
Moving
D(1,-3,1)
Answer:
3/11
Step-by-step explanation:
From the above question, we have the following information
Total number of balls = 12
Number of white balls = 4
Number of blue balls = 3
Number of red balls = 5
We solve this question using combination formula
C(n, r) = nCr = n!/r!(n - r)!
We are told that 3 balls are drawn out at random.
The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!
= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
= 220 ways
The chance of selecting 3 balls at random = 220
To find the chance that all the three balls are selected,
= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls
Chance of selecting (white ball)= 4C1
Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1
Hence,
= [4C1 × 3C1 × 5C1]/ 220
= 60/220
= 6/22
= 3/11
The chance that all three are selected is = 3/11
? = 30 because your number is increasing by -8 how to figure it out is by taking 14 - 6 = 8
