Please help me with this!..
2 answers:
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Answer:
M(x₄ ,y₄) = (-3.5 , 0.5) and
N (x₅ ,y₅) = ( -1 , -0.5 )
Step-by-step explanation:
Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be
M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ
point P( x₁ , y₁) ≡ ( -3 ,3 )
point Q( x₂ , y₂) ≡ (2 , 1 )
point R( x₂ , y₂) ≡ (-4 , -2)
To Find:
M(x₄ ,y₄) = ? and
N (x₅ ,y₅) = ?
Solution:
We have Mid Point Formula as
As M is the mid point of PR and N is the mid point of RQ so we will have
Substituting the given value in above equation we get
∴
∴
Similarly,
∴
∴
∴ M(x₄ ,y₄) = (-3.5 , 0.5) and
N (x₅ ,y₅) = ( -1 , -0.5 )
Answer:
a. N=25
b. X[bar]= 60.52
c. Y[bar]= 106.72
d. SSx= 115.24
e. ∑X*∑Y = 4036684
f. SSxy= 202020.3296
g. √(SSx*SSy)= 449.46
Step-by-step explanation:
Hello!
Using the attached data you need to calculate some statistics.
a) N
The sample size is listed under the first column "subject" You can see that 25 subjects qhere studied so N=25.
b.
The mean of set X is equal to X[bar]= ∑X/n= 1513/25= 60.52
∑X is listed in the second table.
c.
The mean of ser Y is Y[bar]= ∑Y/n= 2668/25= 106.72
∑Y is listed in the second table.
d.
Sum of Squares of set X SSx= ∑X²-[(∑X)²/n]= 91682-[(1513)²/25]= 115.24
e.
∑X*∑Y =1513*2668= 4036684
f.
SSxy= (∑X²-[(∑X)²/n]) * (∑Y²-[(∑Y)²/n])= (91682-[(1513)²/25]) * (286482*[(2668)²/25])= 202020.3296
g.
√(SSx*SSy)= √(115.24*1753)= 449.46
I hope you have a SUPER day!
