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3 years ago

Help With The Two Question

Mathematics

1 answer:

OverLord2011 [107]3 years ago

8 0

For 1st fig:
Angles whose measure is greater than angle 8 is:
<4,<5,<9

For 2nd fig:
Angles whose measure is less than angle 4 is:
<2,<5,<7

▶Explanation:
❇Statement of Exterior angle inequality theorem:
↪An exterior angle of a triangle is greater than either of the non-adjacent interior angles.

➡So for 1st fig ,< 4,<5,<9 were exterior angles of triangle which would obviously be greater angle than interior angle 8 by the theorem.

➡Similarly, for 2nd fig, <2,<5,<7 were non adjacent interior angles which would obviously be less in measure than exterior angle 4.In this case , see both triangles to not get confused.

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Step-by-step explanation:

Given the data in the question;

Sample 1 Sample 2

x"₁ = 259.23 x"₂ = 365.12

s₁ = 34.713 s₂ = 48.297

n₁ = 5 n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

$S_p^2$ = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

$S_p^2$ = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

$S_p^2$ = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [ 4 + 9 ]

$S_p^2$ = [ 4819.9692 + 20993.4 ] / [ 13 ]

$S_p^2$ = 25813.3692 / 13

$S_p^2$ = 1985.64378

Now the Standard Error will be;

$S_{x1-x2$ = √[ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

$S_{x1-x2$ = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

$S_{x1-x2$ = √[ 397.128756 + 198.564378 ]

$S_{x1-x2$ = √595.693134

$S_{x1-x2$ = 24.4068

Critical Value = $t_{\frac{\alpha }{2}, df$ = $t_{0.05, df=13$ = 1.771 { t-table }

So,

Margin of Error E = $t_{\frac{\alpha }{2}, df$ × [ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

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So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666 )

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