Answer: 4-6x^2
Step-by-step explanation:
f•g = f(g(x))
f(g(x)) = 4-2(3x^2) = 4-6x^2
Answer:
24xy6z8
Step-by-step explanation:
Answer:
$12579.84
Step-by-step explanation:
Anna deposited $10,000 into an account three years ago.
The first year she earned 12 percent interest
Interest = 12% of 10000 =![\frac{12}{100} \times 10000=1200](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B100%7D%20%5Ctimes%2010000%3D1200)
Amount at the end of first year = 10000+1200 = 11200
The second year she earned 8 percent interest
Interest = 8% of 11200 =![\frac{8}{100} \times 11200=896](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B100%7D%20%5Ctimes%2011200%3D896)
Amount at the end of second year = 11200+896 = 12096
The third year she earned 4 percent interest.
Interest = 4% of 12096 =![\frac{4}{100} \times 12096=483.84](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B100%7D%20%5Ctimes%2012096%3D483.84)
Amount at the end of third year = 12096+483.84 = 12579.84
Hence she have $12579.84 in her account today
Answer:
P(cell has at least one of the positive nickel-charged options) = 0.83.
P(a cell is not composed of a positive nickel charge greater than +3) = 0.85.
Step-by-step explanation:
It is given that the Nickel Charge Proportions found in the battery are:
0 ==> 0.17
.
+2 ==> 0.35
.
+3 ==> 0.33
.
+4 ==> 0.15.
The numbers associated to the charge are actually the probabilities of the charges because nickel is an element that has multiple oxidation states that is usually found in the above mentioned states.
a) P(cell has at least one of the positive nickel-charged options) = P(a cell has +2 nickel-charged options) + P(a cell has +3 nickel-charged options) + P(a cell has +4 nickel-charged options) = 0.35 + 0.33 + 0.15 = 0.83.
Or:
P(a cell has at least one of the positive nickel-charged options) = 1 - P(a cell has 0 nickel-charged options) = 1 - 0.17 = 0.83.
b) P(a cell is not composed of a positive nickel charge greater than +3) = 1 - P(a cell is composed of a positive nickel charge greater than +3)
= 1 - P(a cell has +4 nickel-charged options) '.' because +4 is only positive nickel charge greater than +3
= 1 - 0.15
= 0.85
To summarize:
P(cell has at least one of the positive nickel-charged options) = 0.83!!!
P(a cell is not composed of a positive nickel charge greater than +3) = 0.85!!!