-x=-y+2
x=y-2
now substitute iin the first one
y=(y-2)^2+5 (y-2)-3
y=y^2-4y+4 +5y-10-3
y=y^2+y-9
y^2-9=0.
y^2=9
y=3
3-x=2
-x=2-3
-x=-1
x=1
the solution are x=1 and y=3
3=1^2+5 (1)-3
3=1+5-3
3=6-3
3=3
Answer:
The area under the normal curve from the mean to 118.8. is 0.47
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
Find the area under the normal curve from the mean to 118.8.
This is the pvalue of Z when X = 118.8 subtracted by the pvalue of Z when X = 100.
X = 118.8
has a pvalue of 0.97
X = 100
has a pvalue of 0.5
0.97 - 0.5 = 0.47
The area under the normal curve from the mean to 118.8. is 0.47
Answer:
it is simplified as much as it possibly can be because there are no like terms. x squared is not the same variable as x.