Equivalent fractions can you help add or subtract because it can be easier to simplify.
hope that helped
Answer:
0.0016283
Step-by-step explanation:
Given that:
Proportion of defective bulbs, p = 30% = 0.3
Sample size, n = 19 bulbs
Probability that the lot will pass inspection :
P(none of the 19 is defective) Or P(only one of the 19 is defective)
P(none of the 19 is defective) = (1 - p) ^n = (1 - 0.3)^19 ; 0.7^19
0.7^19 = 0.0011398
P(only one of the 19 is defective) :
P(1 defective) * P(18 not defective )
(0.3) * (1 - 0.3)^18
0.3 * 0.7^18
0.3 * 0.001628413597910449 = 0.0004885
Hence,
P(none of the 19 is defective) + P(only one of the 19 is defective)
0.0011398 + 0.0004885) = 0.0016283
"He starts both trains at the same time. Train A returns to its starting point every 12 seconds and Train B returns to its starting point every 9 seconds". Basically, what you need to do is find the least common multiple. The least common multiple of 12 and 9 is 36, so the least amount of time, in seconds, that both trains will arrive at the starting points at the same time is 36 seconds.
4.
h(f(x)=h(2x-1)=
(2x-1)^2+1=
4x²-4x+1+1=
4x²-4x+2
5.
f(f(x))=2(2x-1)-1=4x-2-1=4x-3
6.
f o g (x)=f(g(x))
h o g (x)=h(g(x))=
(3x)^2+1=
9x²+1
