Y=x^2-5
y=2x-5
y+5=2x
x=(y+5)/2
y=((y+5)/2)^2-5
y=(y^2+10y+25)/4-5
4y=y^2+10y+25-20
0=y^2+6y+5
0=(y+5)(y+1)
y= -5 or -1
y=2x-5 (y= -5 doesn't work for this one so use -1)
-1=2x-5
2x=4
x=2
Using these answers, you can choose answer D with reasonable certainty.
<span>96-2r=6r-112
Add 2r to both sides
96=8r-112
Add 112 to both sides
208=8r
Divide 8 on both sides
Final Answer: 26=r</span>
Here is your great answer Simplifying
4x2y + 12xy2 + 9y3
Reorder the terms:
12xy2 + 4x2y + 9y3
Factor out the Greatest Common Factor (GCF), 'y'.
y(12xy + 4x2 + 9y2)
Factor a trinomial.
y((2x + 3y)(2x + 3y))
Final result:
y(2x + 3y)(2x + 3y)
Answer:
- Two sided t-test ( d )
- 0.245782 ( c )
- Since P-value is too large we cannot conclude that the students’ weight are different for these two schools. ( c )
- The test is inconclusive; thus we cannot claim that the average weights are different. ( b )
Step-by-step explanation:
1) Test performed is a Two sided test and this because we are trying to determine the mean difference between two groups irrespective of their direction
<u>2) Determine the P-value ( we will use a data-data analysis approach on excel data sheet while assuming Unequal variances )</u>
yes No
Mean 94.47059 89.76471
Variance 173.2647 95.19118
Observations 17 17
df 30
t Stat 1.184211
P(T<=t) one-tail 0.122814
t Critical one-tail 1.697261
P(T<=t) two-tail 0.245782
Hence The p-value = 0.245782
3) Since P-value is too large we cannot conclude that the students’ weight are different for these two schools.
4) The test is inconclusive; thus we cannot claim that the average weights are different.
