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Naya [18.7K]
2 years ago
7

a rectangle has a height of n3+4n2+3n and a width of n^3+5n^2 express the area of the entire rectangle

Mathematics
1 answer:
alukav5142 [94]2 years ago
5 0

<u>Given</u>:

It is given that the height of the rectangle is n^3+4n^2+3n

The width of the rectangle is n^3+5n^2

We need to determine the area of the entire rectangle.

<u>Area of the rectangle:</u>

The area of the rectangle can be determined using the formula,

A=height \times width

Substituting the values, we have;

A=(n^3+4n^2+3n)(n^3+5n^2)

Multiplying each term within the parenthesis, we get;

A=n^{3} n^{3}+n^{3} \cdot 5 n^{2}+4 n^{2} n^{3}+4 n^{2} \cdot 5 n^{2}+3 n n^{3}+3 n \cdot 5 n^{2}

Simplifying, we get;

A=n^{6}+5 n^{5}+4n^{5}+20 n^{4}+3  n^{4}+15 n^{3}

Adding the like terms, we have;

A=n^{6}+9n^{5}+23 n^{4}+15 n^{3}

Thus, the area of the entire rectangle is n^{6}+9 n^{5}+23 n^{4}+15 n^{3}

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Inherently asymmetrical casual relationship.

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The dog owners are given free dog food samples which contain new vegetables. These samples are given to them by organizing booths at the dog events. The reaction of the dog owners is observed towards this new dog food. This an example of inherently asymmetrical relationship.

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Suppose a company borrows 15,000 on 1/1/14 at 10% interest rate for a one year term. The company makes interest payments every q
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A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl
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a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b)   95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

<u>98% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

<u>95% of confidence intervals</u>

({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } }  )

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36     (check table)

X^2_{0.95,13} = 5.892    (check table)

(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13}  } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )

(297.29. (\sqrt{\frac{14-1}{22.36  } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

7 0
2 years ago
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