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Answer with explanation</u>
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Let 
be the average life of light bulbs.
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since 
is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.
Formula for Test statistic : 
where, n= sample size
= sample mean
= Population mean
=population standard deviation
For 
, we have
Using z-value table , Critical one-tailed test value for 0.06 significance level :
Decision : Since critical z value (1.5548) < test statistic (1.6180), so we reject the null hypothesis .
[We reject the null hypothesis when critical value is less than the test statistic value .]
Conclusion : We have enough evidence at 0.06 significance level to support the claim that the new filament yields a longer bulb life
Answer: The dimensions are: " 1.5 mi. × ³⁄₁₀ mi. " .
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{ length = 1.5 mi. ; width = ³⁄₁₀ mi. } .
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Explanation:
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Area of a rectangle:
A = L * w ;
in which: A = Area = (9/20) mi.² ,
L = Length = ?
w = width = (1/5)*L = (L/5) = ?
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A = L * w ; we want to find the dimensions; that is, the values for
"Length (L)" and "width (w)" ;
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Plug in our given values:
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(9/20) mi.² = L * (L/5) ; in which: "w = L/5" ;
→ (9/20) = (L/1) * (L/5) = (L*L)/(1*5) = L² / 5 ;
↔ L² / 5 = 9/20 ;
→ (L² * ? / 5 * ?) = 9/20 ?
→ 20÷5 = 4 ; so; L² *4 = 9 ;
↔ 4 L² = 9 ;
→ Divide EACH side of the equation by "4" ;
→ (4 L²) / 4 = 9/4 ;
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to get: → L² = 9/4 ;
Take the POSITIVE square root of each side of the equation; to isolate "L" on one side of the equation; and to solve for "L" ;
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→ ⁺√(L²) = ⁺√(9/4) ;
→ L = (√9) / (√4) ;
→ L = 3/2 ;
→ w = L/5 = (3/2) ÷ 5 = 3/2 ÷ (5/1) = (3/2) * (1/5) = (3*1)/(2*5) = 3/10;
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Let us check our answers:
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(3/2 mi.) * (3/10 mi.) =? (9/20) mi.² ??
→ (3/2)mi. * (3/10)mi. = (3*3)/(2*10) mi.² = 9/20 mi.² ! Yes!
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So the dimensions are:
Length = (3/2) mi. ; write as: 1.5 mi.
width = ³⁄₁₀ mi.
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or; write as: " 1.5 mi. × ³⁄₁₀ mi. " .
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To much of an open ended question
