Answer:
Two hundred seventy three thousand and fifty .
Step-by-step explanation:
Given : (27 thousands 3 hundreds 5 ones) x 10
Solution:
27 thousands 3 hundreds 5 ones = 27,305
So, 
Periods are counted from last .
The 1st period consists of ones, tens and hundred.
The 2nd period consists of thousand, 10 thousand and 100 thousands.
The 3rd period consists of million, 10 million and 100 million.
So, 273,050 is Two hundred seventy three thousand and fifty
Hence (27 thousands 3 hundreds 5 ones) x 10 is Two hundred seventy three thousand and fifty .
Answer:
akhavkreugvakrehfgv
Step-by-step explanation:
Answer:
square root, 2,12/5,2.47
Step-by-step explanation:
Answer: 197
Step-by-step explanation:
The nth term is 3n+5
so the 64th term will be
(3 x 64)+5
192+5
= 197
Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
