<span>This solution might be a bit out there, but it should probably work. For this, we will consider sequencesto have an invisible "0th" term to make calculation easier.</span>Missing: <span>quadratic</span>
Answer:
Pr(all will survive in one year) = 0.35 to 2 decimal places
Pr(at least 8 will survive) = 0.39 to 2 decimal places.
Step-by-step explanation:
Probability of a tree surviving one year is 0.9. While not surviving probability is 1 - 0.9 = 0.1
This is obtained from the 90℅ surviving rate for one year planted by the landscaping firm.
Probability that if 10 trees are planted, all will survive for one year is:
0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9 = 0.9^10 = 0.349
Probability that at least 8 will survive in one year can be expressed as
8 survived x 2 not survived +(or) 9 survived x 1 not survived +(or) 10 survived
= [(0.9^8)x(0.1^2) + (0.9^9)x(0.1^1) + (0.9^10)] = 0.391
Step-by-step explanation:
Since it is the midpoint we can say:
12x - 1 = 8x +11
All we have do to is get all the x's alone on one side
12x - 1 = 8x + 11
12x = 8x + 12 (add both sides by 1)
4x = 12 (subtract both sides by 8x)
x = 3 (divide both sides to get what x is)
Honestly, the most intelligent solution would be to light the candle first, since it has a longer duration and then I can light the lamp, and then I can light the newspaper. Then, I will use the newspaper to light the fire.
