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USPshnik [31]
3 years ago
6

Employees in the marketing department of a large regional restaurant chain are researching the amount of money that households i

n the region spend in restaurants annually. The employees are looking at how the amount spent annually in restaurants by households led by people ages 25 to 34 compares to households led by people ages 45 to 54. To conduct the research, the employees randomly select 12 households in the region led by people ages 25 to 34 and 12 households in the region led by people ages 45 to 54. Each household is asked how much was spent in restaurants during the previous 12 months. The amounts spent for each household, in dollars, are provided in the samples below. One household in each group spent $1,851 in restaurants during the previous 12 months. Calculate the z-score for the household that spent $1,851 for each group. Use a TI-83, TI-83 plus, and TI-84 calculator.
Round your answers to two decimal places.
25 to 34 45 to 54
1329 2268
1906 1965
2426 1149
1826 1591
1239 1682
1514 1851
1937 1367
1454 2158
Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Answer:

The <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34              45 to 54

  1329                    2268

  1906                    1965

 2426                     1149

  1826                     1591

  1239                    1682

   1514                     1851

  1937                     1367

  1454                    2158

Compute the mean and standard deviation for the group "25 to 34" as follows:

\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01

Compute the <em>z</em>-score for the group "25 to 34" as follows:

z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37

Compute the mean and standard deviation for the group "45 to 54" as follows:

\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39

Compute the <em>z</em>-score for the group "45 to 54" as follows:

z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25

Thus, the <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

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Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

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To find their regular monthly payment, we have:

PV = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}}    \end {bmatrix}

128000 = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

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To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

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PV= 921.433 \begin {bmatrix}  \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

PV = $121718.2714

C) In order to estimate how many payments of $1400  it will take to pay off the loan, we have:

121718.2714 =  \begin {bmatrix}  \dfrac{1300  (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}}    \end {bmatrix}

121718.2714 = 200000  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

\dfrac{121718.2714}{200000 } =  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

0.60859 =  \begin {bmatrix}  (1 - \dfrac{12}{12.078}))^{nt}   \end {bmatrix}

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= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88  - 177795.38

The amount of interest saved = $153920.5

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Explanation:

Given that the length of the tangent AR is 12

The length of SR is 7.7

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Applying the tangent secant theorem, we have,

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Rewriting the above equation, we get,

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