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In-s [12.5K]
3 years ago
12

Bonita spent $8.94 on groceries. She bought a gallon of milk for $4.29 and 3pounds of sliced turkey. How much does 1 pound of sl

iced turkey cost?
Mathematics
1 answer:
Vesna [10]3 years ago
5 0
<h2>Information</h2>

$8.94 spent in total on groceries

Gallon of milk = $4.29

3 pounds of sliced turkey bought = ?


<h2>Explanation</h2>

Subtract the total amount spent on groceries with what items you have prices for (in this case the gallon of milk):

$8.94 (total spent) - $4.29 (gallon of milk) = $4.65 (for 3 pounds of sliced turkey)

Next, divide the price for 3 pounds of sliced turkey with however many pounds Bonita bought (in this case, 3 pounds):

$4.65 (total for 3 pounds of turkey) ÷ 3 (pounds of turkey) = $1.55 (per pound of turkey)


<h2>Solution</h2>

Each pound of turkey costs $1.55

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The volume of the smaller cube is 216 m³.

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(\frac{l}{L})^3 =\frac{216}{2744}

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You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 y
lys-0071 [83]

Answer:

99% confidence interval for the true mean resale value of a​ 5-year-old car of this​ model is [$12,173.24 , $13,306.76].

Step-by-step explanation:

We are given that you manage to obtain data on 17 recently resold​ 5-year-old foreign sedans of the same model.

These 17 cars were resold at an average price of $ 12 comma 740 with a standard deviation of $ 800.

Firstly, the pivotal quantity for 99% confidence interval for the true mean is given by;

                      P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average price = $12,740

            s = sample standard deviation = $800

            n = sample of cars = 17

            \mu = true population mean

<em>Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 99% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.921 < t_1_6 < 2.921) = 0.99  {As the critical value of t at 16 degree of

                                            freedom are -2.921 & 2.921 with P = 0.5%}  

P(-2.921 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.921) = 0.99

P( -2.921 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.921 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X-2.921 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.921 \times {\frac{s}{\sqrt{n} } } ) = 0.99

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Therefore, 99% confidence interval for the true mean resale value of a​ 5-year-old car of this​ model is [$12,173.24 , $13,306.76].

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