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ololo11 [35]
3 years ago
5

Min Jee is renovating a house. The living room is a rectangle 22 2/3 feet long and 17 1/4 feet wide. Min Jee wants to put in new

flooring in the living room. The flooring is sold by the square yard for $13.59. How much flooring does she need and how much will it cost?
Mathematics
2 answers:
krok68 [10]3 years ago
5 0
She will nee 391 square feet of flooring at $13.59: $5313.69
SVEN [57.7K]3 years ago
3 0

Answer

Find out  the  how much flooring does she need and how much will it cost .

To prove

As given

Min Jee is renovating a house.

The\ living\ room\ is\ a\ rectangle\ 22 \frac{2}{3} feet\ long\ and\ 17 \frac{1}{4}\ feet \wide.

i.e

The\ living\ room\ is\ a\ rectangle\ \frac{68}{3} feet\ long\ and\ \frac{69}{4}\ feet \wide.

Formula

Area of a rectangle = Length × Breadth

Put all in the formula

flooring\ Min\ Jee\ needs = \frac{68\times 69}{3\times 4}

flooring\ Min\ Jee\ needs = \frac{4692}{12}

flooring\ Min\ Jee\ needs = 391\ feet^{2}

As 1 feet = 0.3 yard

Flooring Min Jee needs = 391 × 0.3 × 0.3

                                        = 35.19 yard²(Approx)

As given

The flooring is sold by the square yard for $13.59.

Thus

The cost of the flooring = 13.59 × 35.19

                                       = $478.23

Therefore the flooring Mic Jee needs be 35.19 yard²(Approx) and the cost of the flooring is $478.23 .


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3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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3 years ago
<img src="https://tex.z-dn.net/?f=Compare%20%5C%3A%20the%20%5C%3A%20graphs%20%5C%3A%20of%20%5C%5C%205x%20-%20y%20%3D%2012%20%5C%
fenix001 [56]

Answer:

<u>Given</u>

  • Equation 5x - y = 12

and

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<em>See the graphs attached</em>

To draw the graphs follow the rules we described in the previous questions.

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We see the only difference the expressions have is the equation or inequality symbols.

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