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LekaFEV [45]
3 years ago
8

To make a profit, a company’s revenue must be greater than its operating costs. The company’s revenue is modeled by the expressi

on 7.5x – 100, where x represents the number of items sold. The company’s operation costs are modeled by the expression 79.86 + 5.8x. How many items does the company need to sell to make a profit?
The inequality that will determine the number of items that need to be sold to make a profit is
.

The solution to the inequality is
.

The company must sell at least
items to make a profit.
Mathematics
2 answers:
Zina [86]3 years ago
8 0

Answer:

A, D, C

Step-by-step explanation:

ahrayia [7]3 years ago
6 0

Answer:

Inequality: 7.5x – 100 > 79.86 + 5.8x

Solution: x>106

The company must sell at least  106 items to make a profit.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

Revenue: 7.5x – 100

Costs: 79.86 + 5.8x

To make a profit, a company’s revenue must be greater than its operating costs.

So:

7.5x – 100 > 79.86 + 5.8x

Solving for x (number of items sold)

7.5x-5.8x>79.86+100

1.7x >179.86

x>179.86/1.7

x> 105.8

x>106

The company must sell at least 106 items to make a profit.

Feel free to ask for more if needed or if you did not understand something.

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

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$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

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The F-test  statistics value is :

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Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

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Answer:

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