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3 years ago

A vertical parabola goes through (1,-2), (-3, 10), and (4, 31). Use matrices to

Mathematics

1 answer:

Elenna [48]3 years ago

8 0

Answer:

D = 1/2 , E = -1/2 , F = -5/2

Step-by-step explanation:

vertical parabola : ( 1,-2), ( -3, 10 ), (4, 31 )

General form ; $x^2 + y^2 + Dx + Ey + F = 0$

( 1 ,-2 ) = x1, y1

( -3, 10 ) = x2, y2

( 4, 31 ) = x3, y3

hence :

$y1 = ax_{1} ^2 + bx_{1} + c$

$y2 = ax_{2} ^2 + bx_{2} + c$

$y3 = ax_{3} ^2 + bx_{3} + c$

attached below is the detailed solution required

D = 1/2 , E = -1/2 , F = -5/2

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4 years ago

An article reports on an experiment in which each of five groups consisting of six rats was put on a diet with a different carbo

loris [4]

Answer:

H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

Step-by-step explanation:

A B C D E

2.58 2.63 2.13 2.41 2.49

∑A=2.58 ∑B=2.63 ∑C=2.13 ∑D=2.41 ∑E=2.49

A² B² C² D² E²

6.6564 6.9169 4.5369 5.8081 6.2001

∑A²=6.6564 ∑B²=6.9169 ∑C²=4.5369 ∑D²=5.8081 ∑E²=6.2001

Calculations

Group A B C D E Total

N n1=1 n2=1 n3=1 n4=1 n5=1 n=5

∑xi T1=2.58 T2=2.63 T3=2.13 T4=2.41 T5=2.49 ∑x=12.24

∑x²i 6.6564 6.9169 4.5369 5.8081 6.2001 ∑x²=30.1184

Mean xi 2.58 2.63 2.13 2.41 2.49 Total2.448

Σ Σ xij^2 = 185.4 given

Let k = the number of different samples = 5

n=n1+n2+n3+n4+n5=1+1+1+1+1=5

Overall ˉx=12.24/5 =2.448

∑x=T1+T2+T3+T4+T5=2.58+2.63+2.13+2.41+2.49=12.24----1

Correction Factor=(∑x)²/n=12.242/5=29.9635--------(2)

∑T²i/ni=(2.582/1+2.632/1+2.132/1+2.412/1+2.492/1)=30.1184------(3)

∑x²=∑x²1+∑x²2+∑x²3+∑x²4+∑x²5

=6.6564+6.9169+4.5369+5.8081+6.2001=30.1184-------(4)

ANOVA:

Step-1 : sum of squares between samples

SSB=(∑T2i/ni)-(∑x)2/n

=(3)-(2)

=30.1184-29.9635

=0.1549

Step-2 : Total sum of squares

SST=SSB-CF

= 185.4-2.488

=182.952

Step-3 : Sum of Squares Within Samples

SSW=∑x²-(∑T²i/ni)=SST- SSB=

= 182.952-0.1549=182.7971

Step-4 : variance between samples

MSB=SSB/k-1

=0.1549/4

=0.0387

Step-5 : variance within samples

MSW=SSW/n-k

=182.7971/5-5

=36.55942

Step-6 : test statistic F for one way ANOVA test

F=MSB/MSW

=0.0387/36.55942 = 0 .001058

ANOVA table

Source Sums Degrees Mean

of Variation of Squares of freedom Squares

SS DF MS F

B/w samples SSB = 0.1549 k-1 = 4 MSB = 0.0387 0.001

Within samples SSW = 0 n-k = 0 MSW = 36.55942

Total SST = 0.1549 n-1 = 4

H0 : The true average DNA content is not affected by the type of carbohydrate in the diet between samples

Ha : The true average DNA content is affected by the type of carbohydrate in the diet between samples

F(4,0) at 0.05 level of significance =0.5

As calculated F=0.001 >0.5

So, H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

6 0

3 years ago