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Harrizon [31]
3 years ago
7

What is the length of the rectangle?

Mathematics
1 answer:
gregori [183]3 years ago
4 0

Step-by-step explanation:

area \: of \: rectangle \:  = 60 \:  {yards}^{2}  \\  \therefore \: (w + 4) \times \: w = 60\\  \\  \therefore \:  {w}^{2}  + 4w -60 = 0 \\  \\  \therefore \: {w}^{2}  + 10w - 6w -60 = 0 \\  \\   \therefore \: w({w} + 10) - 6(w  + 10)= 0 \\  \\  \therefore \: ({w} + 10) (w- 6)= 0 \\ \therefore \: {w} + 10 = 0 \: or \: w- 6 = 0 \\  \therefore \: {w}   = -  10  \: or \: w =  6  \\  \because \: sides \: of \: rectangle \: cant \: be \: negative \\  \therefore \: w \neq \:    - 10 \\  \\ \therefore \: w = 6 \\  \\ \therefore \: w + 4 = 6 + 4 = 10 \\  \\ length \: of \: rectangle \:  = 10 \: yards.

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Step-by-step explanation:

Mean = 81740

Standard deviation = 4590

Sample size = 15

Alpha level = 1-0.95 = 0.05

Df = 15-1 = 14

Critical value:

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t0.025

t critical value = 2.145

Margin of error ME

2.145 x 4590/√15

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ME = 2542.09

Confidence interval

Lower CI = mean - ME

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= 79197.91

Upper CI = mean + ME

= 81740+2542.09

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[ 79197.91, 84282.09]

B.

Using excel, exact answer for CI

Lower limit = 79198.142724212173

Upper limit = 84281.8572757827

C.

The assumptions to be made from the population are that

1. Samples are random

2. These samples are gotten from an approximately normal distribution

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