Answer:
PΔJKL=66
Step-by-step explanation:
so we are given the line segments JK, KL, and LJ which are tangent to k(O), and also that JA=9, AL=10, and CK=14
JL=JA+AL (parts whole postulate)
JL=9+10=19 (substitution, algebra)
JA=JB=9 (tangent segments from the same point are congruent)
CK=KB=14 (tangent segments from the same point are congruent)
JK=JB+KB (parts whole postulate)
JK=9+14=23 (substitution, algebra)
LA=LC=10 (tangent segments from the same point are congruent)
LK=LC+CK (parts whole postulate)
LK=10+14=24 (substitution, algebra)
Perimeter of ΔJKL=LK+KL+LJ (perimeter formula for triangles)
Perimeter of ΔJKL=23+24+19=66 (substitution, algebra)
Answer:
Step-by-step explanation:
Given that a manufacturer makes three types of screws
Type A B C Total
Pack 1000 500 800
P for flaw 0.001 0.03 0.005
Pack*p 1 15 4 20
a) Expected total number of defective screws = 20
b) For one pack each no of defective screws =20
Hence for 5 expected number number of packs form each should be 1/3
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