A real-estate company appraised the market value of 36 homes in Lyttelton and found that the sample mean and standard deviation were $150,000 and $17,000 respectively. The real-estate company also appraised the market value of 45 homes in Aranui and found that the sample mean and standard deviation were $100,000 and $12,000 respectively. Calculate the 90% confidence interval estimate for the population difference in market value between the Lyttelton and Aranui areas .

Answers

Answer 1

Solution :

                    Sample size       Sample mean             Sample S.D.

Sample 1        [tex]$n_1=36$[/tex]              [tex]$\bar{x}_1=150,000$[/tex]               [tex]$s_1=17,000$[/tex]

Sample 2       [tex]$n_2=45$[/tex]              [tex]$\bar{x}_2=100,000$[/tex]               [tex]$s_2=12,000$[/tex]

[tex]$df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}$[/tex]

   = 60

Therefore, significance level, α = 0.1

Critical value, t* = 1.6706

So, the margin of error, [tex]$t^*=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$[/tex]

                                          = 559.9896

Lower limit, [tex]$(\bar x_1-\bar x_2)-\text{(margin of error)}=44402.0104$[/tex]

Upper limit,  [tex]$(\bar x_1-\bar x_2)+\text{(margin of error)}=55597.9896$[/tex]

Therefore 90% C.I. is (44402.0104, 55597.9896)   or [tex]$44402.0104 < \mu_1 - \mu_2 < 55597.9896$[/tex]


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Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Answers

Answer:

The equivalent will be:

[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex]

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}[/tex]

Let us solve the expression step by step to get the equivalent

[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}[/tex]

as

[tex]\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}[/tex]      ∵ [tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]

[tex]=x^{2\cdot \frac{1}{7}}[/tex]

[tex]=x^{\frac{2}{7}}[/tex]

also

[tex]\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}[/tex]         ∵  [tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]

[tex]=y^{3\cdot \frac{1}{5}}[/tex]

[tex]=y^{\frac{3}{5}}[/tex]

so the expression becomes

[tex]\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}[/tex]

[tex]=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex]            ∵ [tex]\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}[/tex]

Thus, the equivalent will be:

[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex]

Therefore, option 'a' is true.

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Answer:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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determine the domain and range of the function f(x)= -|x| + 2.
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Answers

Answer:

For the given function f(x)= -|x| + 2

the domain of the function is: All real numbers

the range of the function is: all numbers greater than or equal to 2

Step-by-step explanation:

We need to find the domain and range of the function f(x)= -|x| + 2.

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Answers

Answer:

Step-by-step explanation:

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