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Alexxx [7]
2 years ago
15

Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown.

Mathematics
1 answer:
shepuryov [24]2 years ago
7 0

The smallest possible whole-number length of the unknown side is 17 inches.

<h3>What is the Pythagoras theorem?</h3>

The Pythagoras theorem states that the square of the longest side must be equal to the sum of the square of the other two sides in a right-angle triangle.

From the information given, the sides of an obtuse triangle measure 9 inches and 14 inches.

Therefore, the third side will be:

c² = 9² + 14²

c² = 81 + 196

c² = 277

c = ✓277

c = 16.64

c = 17

Hence, the smallest possible whole-number length of the unknown side is 17 inches.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ1

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he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w
stellarik [79]

Answer:

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

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

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

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   )

3 0
3 years ago
What is the solution to the linear equation?
gizmo_the_mogwai [7]

Answer:

p = 30/1/3 or 91/3 or 30.33

Step-by-step explanation:

2 + p = 7/3 + 30

p = 32/1/3 - 2

p = 30/1/3

3 0
3 years ago
Simplify (6^5/7^3)^2
o-na [289]
Simplify the following:
(6^5/7^3)^2

7^3 = 7×7^2:
(6^5/(7×7^2))^2

7^2 = 49:
(6^5/(7×49))^2

7×49 = 343:
(6^5/343)^2

6^5 = 6×6^4 = 6 (6^2)^2:
((6 (6^2)^2)/343)^2

6^2 = 36:
((6×36^2)/343)^2

 | | 3 | 6
× | | 3 | 6
 | 2 | 1 | 6
1 | 0 | 8 | 0
1 | 2 | 9 | 6:
((6×1296)/343)^2

6×1296 = 7776:
(7776/343)^2

(7776/343)^2 = 7776^2/343^2:
7776^2/343^2

 | | | | 7 | 7 | 7 | 6
× | | | | 7 | 7 | 7 | 6
 | | | 4 | 6 | 6 | 5 | 6
 | | 5 | 4 | 4 | 3 | 2 | 0
 | 5 | 4 | 4 | 3 | 2 | 0 | 0
5 | 4 | 4 | 3 | 2 | 0 | 0 | 0
6 | 0 | 4 | 6 | 6 | 1 | 7 | 6:
60466176/343^2

 | | | 3 | 4 | 3
× | | | 3 | 4 | 3
 | | 1 | 0 | 2 | 9
 | 1 | 3 | 7 | 2 | 0
1 | 0 | 2 | 9 | 0 | 0
1 | 1 | 7 | 6 | 4 | 9:

Answer: 60466176/117649
6 0
3 years ago
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