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Shalnov [3]
3 years ago
6

The owner of ProPhone has charted the company's marginal revenue and marginal cost for its latest line of smartphones, the Blaze

r. Use the chart to calculate the company's profit. How many phones will ProPhone need to sell to maximize profit?

Mathematics
2 answers:
steposvetlana [31]3 years ago
5 0
Profit is total revenue minus total cost. So, the profit for each row of data is:
Quantity | Profit
0 | 30 (loss)
1 | 30
2 | 160
3 | 190
4 | 260
5 | 300
6 | 310
7 | 290

The company has to produce 6 quantities to maximize profit.
lakkis [162]3 years ago
3 0

Answer:

6 phones

Step-by-step explanation:

To calculate the profit, the total revenue is subtracted by total cost.

We will calculate the company's profit as per given chart.

Quantity of phones     Total revenue-total cost             Profit/loss

      0                                              0 - 90                              -90  (loss)

      1                                           200 - 170                             30

      2                                          360 - 200                           160

      3                                          480 - 290                           190

      4                                          600 - 340                           260

      5                                          700 - 400                           300

      6                                          810 - 500                            310

      7                                          910 - 620                            290

The profit for selling 6 phone units is 310 which is highest.

Therefore ProPhone  will need to sell six phones to maximize profit.

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4 0
2 years ago
The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ou
Setler [38]

Answer:

Step-by-step explanation:

(a)

The distribution of X is Normal Distribution with mean = \mu =17 and Variance = \sigma^{2} = 16 \ i.e., X \sim N (17, 16),

(b)

The distribution of \bar{x} is Normal Distribution with mean = \mu =17 and Variance = \sigma^{2}/n = 16/15= 1.0667.i.e., \bar{x}\sim N(17,1.0667)

c)

To find P(15.5 < X < 18):

Case 1: For X from 15.5 to mid value:

Z = (15.5 - 17)/4 = - 0.375

Table of Area Under Standard Normal Curve gives area = 0.1480

Case 2: For X from mid value to 18:

Z = (18 - 17)/4 = 0.25

Table of Area Under Standard Normal Curve gives area = 0.0987

So,

P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467

So,

Answer is:

0.2467

(d)

SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}

= 1.0328

To find P(15.5 < \bar{x}< 18):

Case 1: For \bar{x} from 15.5 to mid value:

Z = (15.5 - 17)/1.0328 = - 1.4524

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 18:

Z = (18 - 17)/1.0328 = 0.9682

Table of Area Under Standard Normal Curve gives area = 0.3340

So,

P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605

So,

Answer is:

0.7605

(e)

Correct option:

No

because Population SD is provided.

(f)

(i)

Q1 is given by:

- 0.6745 = (\bar{x} - 17)/1.0328

So,

X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034

So,

Q1 = 16.3034

(ii)

Q3 is given by:

0.6745 = (\bar{x} - 17)/1.0328

So,

X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966

So,

Q3= 17.6966

(iii)

IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932

So

Answers are:

Q1 = 16.3034 ounces

Q3 = 17.6966 Ounces

IQR = 1.3932 Ounces

8 0
3 years ago
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