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aev [14]
3 years ago
7

Jem boy wants to make his 8-meter square pool into a rectangular one by increasing its length by 2m and decreasing its width by

2m .jem boy asked your expertise to help him decide on certain matters.
a.What will be the new dimensions of jem boy's pool?
b.What will be the new are of Jem Boy's pool? What special product will be used?
c.If the sides of the square pool is unknown,how will you represent its area?
d. I fJem Boy does not want the area of his pool to decrease, will he pursue his plan? Explain your answer.
Mathematics
1 answer:
butalik [34]3 years ago
3 0
Side = 8m
He increases length by 2m and reduces width by 2m
(a.) New dimensions will be -
Length = 10m and Width = 6m
(b.) Area = length x breadth (or width)
 here we will use the special product (x+y) (x-y)
length = (x+y) and width = (x-y)
area = (8+2) x (8-2)
            8^(2) - 2^(2) 
            = 60
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Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval in which we are a percentage sure the true mean is in the interval.

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x_{1}-x_{2} ± t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }

where

x₁ and x₂ are sample means

t is t-score

S_{p} is estimate of standard deviation

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The estimate of standard deviation is calculated as

S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

df=n_{1}+n_{2}-2

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }

S_{p}=\sqrt{\frac{299.07}{19} }

S_{p}= 15.74

The difference between means is

x_{1}-x_{2} = 17.5 - 13.66 = 3.84

Calculating the interval:

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{\frac{21}{108} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{0.194}

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.

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3 years ago
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