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Sidana [21]
3 years ago
12

You are in charge of purchases at the student-run used-book supply program at your college, and you must decide how many introdu

ctory calculus, history, and marketing texts should be purchased from students for resale. Due to budget limitations, you cannot purchase more than 700 of these textbooks each semester. There are also shelf-space limitations: Calculus texts occupy 2 units of shelf space each, history books 1 unit each, and marketing texts 4 units each, and you can spare at most 1,200 units of shelf space for the texts. If the used book program makes a profit of $10 on each calculus text, $4 on each history text, and $8 on each marketing text, how many of each type of text should you purchase to maximize profit? HINT [See Example 3.]
calculus text(s) =

history text(s) =

marketing text(s) =

What is the maximum profit the program can make in a semester?
Mathematics
1 answer:
mariarad [96]3 years ago
7 0

Answer:

  • Calculus texts: 600
  • History texts: 0
  • Marketing texts: 0

Step-by-step explanation:

Each Calculus text returns $10/2 = $5 per unit of shelf space. For History and Marketing texts, the respective numbers are $4/1 = $4 per unit, and $8/4 = $2 per unit. Using 1200 units of shelf space for 600 Calculus texts returns ...

  $5/unit × 1200 units = $6000 . . . profit

Any other use of units of shelf space will reduce profit.

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A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \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 sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0
3 years ago
Fas
Mrrafil [7]

The graph is vertically stretched by a factor of 2 and translated 3 units right when it is transformed. Option A is correct.

<h3>What is transformation of a function?</h3>

Transformation of a function is shifting the function from its original place in the graph.

Types of transformation-

  • Horizontal shift- Let the parent function is f(x). Thus by replacing parent function with f(x-b) shifts the graph b units right and by replacing parent function with f(x+b) shifts the graph b units left.
  • Vertical shift- Let the parent function is f(x). Thus by replacing parent function with f(x)-c shifts the graph c units down and by replacing parent function with f(x)+c shifts the graph c units up.

The given function is,

f(x) = x^3

This function is changed to the function,

g(x) = 2f(x -3),

Here the 3 units is substrate in the function. Thus, it is shiftet 3 units right. The number 2 is multiplied in the function which vertically stretched the graph by a factor of 2.

Thus, the graph is vertically stretched by a factor of 2 and translated 3 units right when it is transformed. Option A is correct.

Learn more about the transformation of a function here;

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