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erik [133]
3 years ago
7

A factory sells backpacks for $40.00 each. The cost to make 1 backpack is $10.00. In addition to the cost of making backpacks, t

he factory has operating expenses of $12,000 per week. The factory's goal is to make a profit of at least $9,800 per week. Which inequality represents the number of backpacks, x, that need to be sold each week for the factory to meet this goal?
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
7 0

Answer:

<h2>The factory needs to sell 327 packbacks to make at least 9,800 per week.</h2>

Step-by-step explanation:

We know that each backpacks is sold for $40.00.

The goal is to make at least $9,800 per week. With this information we can define the inequality

40x -10x \leq 9,800

Where x represents backpacks. Notice that this inequality is about profits, that's why we subtract the cost from the sell price, in this case, the profid margin is $30.00 per backpack, so

30x\leq 9,800

Solving for x

x\leq \frac{9800}{30}\\ x \leq 326.67

Therefore, the factory needs to sell 327 packbacks to make at least 9,800 per week.

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x=1/4

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Step-by-step explanation:

4x -y =3

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substitute y =-8x into the first equation.  Everywhere you see y put 8x

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8 0
3 years ago
Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

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                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

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