Okay guys. I have to take the Algebra 2 EOC tomorrow, and I have a couple questions. My algebra teacher doesn't teach very well at all.. So I'm going to post a few questions and see if anyone can explain them to me.
1) What are all the roots of
x^3 - 3x^2 - x + 3 ?
I thought it would be 1, and -1, but they threw in a 3 and I have no idea why. ??
2) The mileage of the vehicles in a government fleet are normally distributed with a mean of 60,000 miles and a standard deviation of 8,000 miles. If a vehicle from the fleet is randomly selected, which is the closest to the probability that the mileage is greater than 44,000 miles?
That one I have no clue about...
These are just a few. If these can be done on the TI-84 plus calculator, I have one... :)
Please help if you can!
This EOC counts as my final exam grade!!
Thanks bunches!!
1) What are all the roots of
x^3 - 3x^2 - x + 3 ?
I thought it would be 1, and -1, but they threw in a 3 and I have no idea why. ??
2) The mileage of the vehicles in a government fleet are normally distributed with a mean of 60,000 miles and a standard deviation of 8,000 miles. If a vehicle from the fleet is randomly selected, which is the closest to the probability that the mileage is greater than 44,000 miles?
That one I have no clue about...
These are just a few. If these can be done on the TI-84 plus calculator, I have one... :)
Please help if you can!
This EOC counts as my final exam grade!!
Thanks bunches!!
-
1) Were you provided with the answers 1, -1, and 3? If so, then the reason why 3 is a root is because if you plug it in for x, the expression evaluates to 0, the definition of a 'root'. If you weren't provided with those solutions, then that means you had to solve it, which isn't a very easy task. The formula that solve for cubic roots is VERY long: http://en.wikipedia.org/wiki/Cubic_funct… (thought the quadratic formula was bad? Think again!)
2) This is a simple statistics question. Why it's on an Algebra 2 test, I have no idea. Either way, it's not too difficult to solve. In fact, since you have a TI-84 Plus calculator, you can solve it with one command: [2nd][Vars](Distr) > 2:normalcdf(.
normalcdf(44000,10^99,60000,8000) ≈ .977 ≈ 98%
If you want to do it the long way, you'll have to calculate the z-score, the formula of which is:
z = (x − µ)/σ = (44000 − 60000)/8000 = -2
Now that you know the z-score is -2, you can do normalcdf(-2,10^99) ≈ .977.
2) This is a simple statistics question. Why it's on an Algebra 2 test, I have no idea. Either way, it's not too difficult to solve. In fact, since you have a TI-84 Plus calculator, you can solve it with one command: [2nd][Vars](Distr) > 2:normalcdf(.
normalcdf(44000,10^99,60000,8000) ≈ .977 ≈ 98%
If you want to do it the long way, you'll have to calculate the z-score, the formula of which is:
z = (x − µ)/σ = (44000 − 60000)/8000 = -2
Now that you know the z-score is -2, you can do normalcdf(-2,10^99) ≈ .977.
-
10^99 just represents a really big number. If we could type infinity in our calculators, that would work too. Using a really high number ensures that we cover all the area under the Normal curve. If you needed to find the probability that mileage is LESS than 44000, you would go from -10^99 to -2.
Report Abuse
-
Im in Algebra 1 now so im using what i know, im not sure if this would help though... for 1) itd have the 3 bc its a root of the exponent, 2) im with you no clue but i know both can be done on a calculator, couple websites just search it up.... this probablly doesnt help worth a shot good luck