Given A={1,2,3}, B={2,4,6} and C={1,2,3,4,5,6}, then A ∩ (B ∩ C) =
oksian1 [2.3K]
Answer:
Step-by-step explanation:
A={1,2,3}
B={2,4,6}
C={1,2,3,4,5,6}
B∩C={2,4,6}
A∩(B∩C)={1,2,3}∩{2,4,6}={2}
A+b=15----(1)
a*b=54----(2)
(1) a=15-b ----(3)
put(3) in(2)
(15-b)*b=54
15b-b^2=54
b^2-15b+54=0
(b-9)(b-6)=0
so b=9 or b=6
if b=9 then a=15-9=6
if b=6 then a=15-6=9
answer two numbers are 6 and 9
Answer:
0.242
Step-by-step explanation:
p = 0.08, q = 1 − p = 0.92, and n = 1000.
The mean and standard deviation are:
μ = p = 0.08
σ = √(pq/n) = 0.00858
The z score is:
z = (x − μ) / σ = -0.70
Using a calculator or z score table, the probability is:
P(Z < -0.70) = 0.242
9×10⁴ is 300 times greater than 3×10²
as, (9×10⁴)/(3×10²)=3× 10²
and the proper of writting powers in computer form is a×10^b
Answer:
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