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WARRIOR [948]
2 years ago
10

The production cost, represented by y, of a Bluetooth device, is equal to the product of labour cost (m) and number of devices (

x) in addition to a constant equipment renting fee (q) 1.1 write an equation to represent the production cost in terms of m, x and q. 1.2 Change the subject of the formula in 1.1 to: 1.2.1 m; 1.2.2 q, 1.2.3 x. Determine the cost if labour amounts to R1,245, with an equipment renting fee of R250, to produce 5 devices. 1.4 Suppose the production cost is R 18,925, labour cost is and equipment renting fee is R250, determine the number of devices produced. 1.5 Hence, write down the equation with y as the subject (in terms of x). The value of m and q must e indicated. Discuss whether it is possible that any devices are produced if the production cost is R0,00. 1.7 It is required that the production cost must be less than R123 505. Determine the possible number of devices that will be produced. Show your solution graphically. Use interval notation and indicate the restriction on the number type of your answer. Interpret your answer
Mathematics
1 answer:
pogonyaev2 years ago
5 0

It should be noted that the equation to represent the production cost will be y = mx + q.

<h3>How to calculate the equation</h3>

From the information given, the production cost, represented by y, of a Bluetooth device, is equal to the product of labour cost (m) and number of devices (x) in addition to a constant equipment renting fee (q). This will be: y = mx + q.

Changing the subject of the formula to m will be:

y = mx + q

y - q = mx

m = (y - q)/x

Learn more about equations on:

brainly.com/question/1214333

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ICE Princess25 [194]

Answer:

a. Error is 0.14

b. between 1913.68 to 1772.32

Step-by-step explanation:

Requirement a)

The margin of error is a statistic expressing the amount of random sampling error in a survey's results.

Given,

Size of the sample n = 50

We know, At 95% confidence interval,

The margin of error,

= 0.98/√n  

= 0.98/√50

= 0.98/7.07

= 0.13859 ~ 0.14

So the margin of error at 95% confidence level is 0.14.

Requirement b)

Given,

Standard deviation σ = 255

Population mean z = 1814

Size of the sample n = 50

We know, At 95% confidence interval,

(z-µ)/(σ/√n)  < Z_0.05

= - Z_0.05 < (z-µ)/(σ/√n) < Z_0.05

= - 1.96 * σ/√n < (z-µ)/(σ/√n)*  σ/√n  < 1.96 * σ/√n  [ in two tail Z test the value of Z_0.05 is 1.96 ]

= - 1.96 * 255/√50 < Z-µ <  1.96 * 255/√50

= -70.6828 < Z-µ < 70.6828

= 70.6828 > µ - Z > -70.6828

= 70.6828 + Z > µ > -70.6828 + Z

= 70.6828 + 1814 > µ > -70.6828 + 1814

=1913.68 > µ > 1772.32

So, the sample mean would be between 1913.68 to 1772.32.

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