Answer.
Answer is 9
Step-by-step explanation:
Multiply height by with.
1,1x8,1
then do whatever that equals times 2,5 and you should get your answer.
You must first normalize this in order to use a Z-score table. To convert what you have into a Z score you must use this formula: (x- mean) /standard deviation. In your case (334-310)/12=2. So now you want the probability that a score is greater than 334, which in turn means you want P(Z>2)=1-P(Z<2)=1-.9772=0.0228=2.28%.
Let k be the scale factor relating two similar prisms

and

, such that for corresponding parts of prisms

and

(for heights, in particular) we have

. In our case

.
For surfaces area we have

.
So, the right answer is 4:25 (choice B)
Answer:
∴Given Δ ABC is not a right-angle triangle
a= AB = √45 = 3√5
b = BC = 12
c = AC = √45 = 3√5
Step-by-step explanation:
Given vertices are A(3,3) and B(6,9)

AB = 
Given vertices are B(6,9) and C( 6,-3)
= 
BC = 12
Given vertices are A(3,3) and C( 6,-3)

AC² = AB²+BC²
45 = 45+144
45 ≠ 189
∴Given Δ ABC is not a right angle triangle