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natka813 [3]
2 years ago
8

Anita is a door-to-door saleswoman for a company selling bedcovers. The company pays her a fixed salary of Rs W every month and

also a bonus of Rs Z for every bedcover sold by her.
In March 2002, Anita sold 46 bedcovers. How much would her total earnings for March be?
a) (W + Z) × 46
b) W + Z × 46
c) (W + 46) × Z
d) W × Z × 46
Mathematics
1 answer:
lorasvet [3.4K]2 years ago
3 0

Given:

Anita's fixed salary = Rs. W

Bonus = Rs. Z for every bedcover sold by her.

To find:

Her total earnings if she sold 46 bedcovers.

Solution:

We have,

Bonus if she sold 1 bedcover = Rs. Z

Bonus if she sold 46 bedcover = Rs. Z × 46

We know that,

Total earnings = Fixed salary + Bonus

                        = W+Z\times 46

Therefore, the correct option is (b).

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nalin [4]

Answer:

1) Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959

Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909

2) Cost/die_1 = \frac{12}{84 x 0.959}=0.149

Cost/die_2 = \frac{15}{100 x 0.909}=0.165

3) Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2

Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2

And for the new yield we need to take in count the increase of 15% for the area and we got this:

Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957

Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905

4) DR_{old}=\frac{1}{\sqrt{0.92}} -1=0.0426 defects/cm^2

DR_{new}=\frac{1}{\sqrt{0.95}} -1=0.0260defects/cm^2

Step-by-step explanation:

Part 1

For this part first we need to find the die areas with the following formula:

Area= \frac{W area}{Number count}

Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2

Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2

Now we can use the yield equation given by:

Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}

And replacing we got:

Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959

Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909

Part 2

For this part we can use the formula for cost per die like this:

Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}

And replacing we got:

Cost/die_1 = \frac{12}{84 x 0.959}=0.149

Cost/die_2 = \frac{15}{100 x 0.909}=0.165

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2

Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2

And for the new yield we need to take in count the increase of 15% for the area and we got this:

Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957

Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}

And if we solve for the Defect rate we got:

DR= \frac{1}{\sqrt{Yield}}-1

Now we can find the previous and new defect rate like this:

DR_{old}=\frac{1}{\sqrt{0.92}} -1=0.0426 defects/cm^2

And for the new defect rate we got:

DR_{new}=\frac{1}{\sqrt{0.95}} -1=0.0260defects/cm^2

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