Triangles ABC and LBM are similar. We know this because AL and LB have the same length, so that AB is twice as long as either AL or LB. The same goes for MC and BM, and BC. The angle B is the same for both tirangles ABC and LBM, so the side-angle-side postulate tells us the triangles are similar, and in particular that triangle ABC is twice as large as LBM.
All this to say that LM must be half as long as AC, so LM has length (B) 14 cm.
Answer:
c)0.65
Step-by-step explanation:
= -30a^5b + 12a^4b^3 + 6a^3b^2 + 10a^3b^2 - 4a^2b^4 - 2ab^3
= -30a^5b + 12a^4b^3 + 16a^3b^2 - 4a^2b^4 - 2ab^3
answer
B. -30a^5b + 12a^4b^3 + 16a^3b^2 - 4a^2b^4 - 2ab^3
Answer:
a) <em>Z-score = 0.75</em>
b) <em>Z-score = -32.833</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that mean of the Population = 33
Given a standard deviation of the Population = 12
Let 'X' be a random variable in a normal distribution
Let 'X' = 42
<u><em>Step(ii):-</em></u>
<em> </em>
<em></em>
<em> </em>
<em></em>
<u><em>Step(iii):-</em></u>
<em>Given that mean of the Population = 89</em>
Given a standard deviation of the Population = 1
Let 'X' be a random variable in a normal distribution
Let 'X⁻ = 82
<em> </em>
<em></em>
<em> </em>
<em></em>
<em>Z-score = -32.833</em>
<em></em>
Answer:

Step-by-step explanation:
GIVEN: A bag contain
red marbles,
green marbles and
blue marble. A marble is chosen at random from the bag and not replaced, then a second marble is chose.
TO FIND: What is the probability both marbles are green.
SOLUTION:
Total marbles in bag 
total number of green marbles 
Probability that first marble will be green 

Probability that second marble will be green 

As both events are disjoint
probability both marbles are green 

Hence the probability both marbles are green is 