Answer:(6, -1)
Step-by-step explanation:
Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that 
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.




IQ scores of at least 130.81 are identified with the upper 2%.
Answer:

Step-by-step explanation:
Step 1: Define
Difference Quotient: 
f(x) = -x² - 3x + 1
f(x + h) means that x = (x + h)
f(x) is just the normal function
Step 2: Find difference quotient
- <u>Substitute:</u>
![\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%2Bh%29%5E2-3%28x%2Bh%29%2B1%5D-%28-x%5E2-3x%2B1%29%7D%7Bh%7D)
- <u>Expand and Distribute:</u>
![\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%5E2%2B2hx%2Bh%5E2%29-3x-3h%2B1%5D%2Bx%5E2%2B3x-1%7D%7Bh%7D)
- <u>Distribute:</u>

- <u>Combine like terms:</u>

- <u>Factor out </u><em><u>h</u></em><u>:</u>

- <u>Simplify:</u>

Answer:
x=1
Step-by-step explanation:
you would change it to 5x-4(-3x+5)=-3, because you put the info for the first equation into the second equation.
5x-4(-3x+5)=-3
5x+12x-20=-3
17x-20=-3
17x=17
x=1