Triangle A E C is shown. Line segment B D is drawn near point C to form triangle B D C.
1 answer:
∠BDC and ∠AED are right angles, is a piece of additional information is appropriate to prove △ CEA ~ △ CDB
Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.
Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.
Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)
Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.
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Answer:
Step-by-step explanation:
given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
Clerk I II III total
Pack 0.33 0.23 0.44 1
Defective 0.02 0.025 0.015
Pack&def 0.0066 0.00575 0.0066 0.01895
a) probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate
= P(II clerk) -P(II clerk and defective) =
b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)
=0.01895
c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3
=P(clerk 3 and def)/P(defective)
=