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Helen [10]
4 years ago
12

05*10^-3 m in long form

Mathematics
1 answer:
butalik [34]4 years ago
5 0
-1000 is the answer to 5*10^-3
You might be interested in
Estion 4
Sunny_sXe [5.5K]

Answer:

Check Explanation

Step-by-step explanation:

Sale price = GHS80 per unit from first week of December to first week of January.

And at a reduced price of 30% from second week of January to the last week of January.

So, sales price for the second period = 70% × 80 = GHS56

To now find the profits for each of the purchase alternatives, we need to calculate the expected total demand

Expected demand units = (Demand × Probability)

First Period

Demand Probability | Expected demand units

500 0.1 | 50

600 0.3 | 180

750 0.4 | 300

850 0.2 | 170

Second period

Demand Probability | Expected demand units

320 0.5 | 160

180 0.3 | 54

130 0.2 | 26

Total expected demand units for first period = 50 + 180 + 300 + 170 = 700

Total expected demand units for second period = 160 + 54 + 26 = 240

i) When a pack of 600 products only is ordered, it is evident that it will cater for only the first period.

Expected Profit = (Expected sales it can cater for) - (Price of one pack of 600 products)

Expected sales it can cater for = 600 × 80 = GHS 48,000

Expected price of one pack of 600 products = 600 × 60 = GHS 36,000

Expected profit = 48000 - 36000 = GHS 12,000

ii) When a pack of 800 products only is ordered, it is evident that it will cater for the entire first period (700) and 100 from the second period.

Expected Profit = (Expected sales it can cater for) - (Price of one pack of 800 products)

Expected sales it can cater for = (700 × 80) + (100 × 56) = 56,000 + 5,600 = GHS 61,600

Expected price of one pack of 800 products = 800 × 57 = GHS 45,600

Expected profit = 61600 - 45600 = GHS 16,000

iii) When a pack of 1000 products only is ordered, it is evident that it will cater for the entire period, 700 and 240.

Expected Profit = (Expected sales it can cater for) - (Price of one pack of 1000 products)

Expected sales it can cater for = (700 × 80) + (240 × 56) = 56,000 + 13,440 = GHS 69,440

Expected price of one pack of 100 products = 1000 × 52 = GHS 52,000

Expected profit = 69440 - 52000 = GHS 17,440

iv) To do this, we first assume that

- the probabilities provided are very correct.

- the products are sold on a first come first serve basis

- the profits per unit for each period is calculated too.

Profit per product in this case = (16000/800) = GHS 20

For the first period

Expected profit = (700 × 80) - (700 × 57) = GHS 16,100

Average profit per unit = (16100/700) = GHS 23

For the second period

Expected profit = (100 × 56) - (100 × 57) = - GHS 100

Average profit per unit = (-100/100) = -GHS 1

Standard deviation = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = [700 × (23-20)²] + [100 × (-1-20)²]

= 6300 + 44,100 = 50,400

N = 800

Standard deviation per unit = √(50400/800) = GHS 7.94

Variance per unit = (standard deviation per unit)² = (7.94)² = 63.

Variance on 800 units = 800 (1² × 63) = 800 × 63 = 50,400

Standard deviation on profits of 800 units = √(50400) = GHS 224.5

v) With the same assumptions as in (iv), but now, we include the Profit (or more appropriately, the loss from unsold units of products)

Profit per product in this case = (17440/1000) = GHS 17.44

For the first period

Expected profit = (700 × 80) - (700 × 52) = GHS 19,600

Average profit per unit = (19600/700) = GHS 28

For the second period

Expected profit = (240 × 56) - (240 × 52) = - GHS 960

Average profit per unit = (960/240) = GHS 4

The expected unsold products = 1000 - 940 = 60

Profit on those unsold products = 0 - (60 × 52) = -GHS 3,120

Profit per unit = (-3120/60) = - GHS 52

Standard deviation = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = [700 × (28-17.44)²] + [240 × (4-17.44)²] + [60 × (-52-17.44)²]

= 78,059.52 + 43,352.064 + 289,314.816 = 410,726.4

N = 1000

Standard deviation per unit = √(410,726.4/1000) = GHS 20.27

Variance per unit = (standard deviation per unit)² = (20.27)² = 410.7264

Variance on 1000 units = 1000 (1² × 410.7264) = 800 × 410.7264 = 410,726.4

Standard deviation on profits of 1000 units = √(410,726.4) = GHS 640.88

vi) The standard deviation on profits show how much the real profits can range below or abobe the expected profit. That is, the standard deviation basically represents how big the risks or rewards can get.

A larger standard deviation will indicate a higher risk in case of loss and a higher reward in case of profits.

The option with the lower risk is the option with the lower standard deviation.

Hence, a pack of 800 products should be ordered instead of a pack of 1000 products as it has a lower standard deviation and hence, a lower risk attached to it thereby minimizing the risk.

Hope this Helps!!!

5 0
4 years ago
Please someone help me<br><br> Only the ODD questions please.
Klio2033 [76]
I hope you can read this, and I hope you understand how I did 7. I probably overworked it, and it was possibly a simpler way to get it.

7 0
3 years ago
there was a total of $548 collected for the school play. the adult tickets cost $6 and the student tickets cost $4. if 12 more s
balandron [24]
You can first make the equations:

6x+4y=548
y-12=x

Here, you can use the substitution method and you get:

6(y-12)+4y=548
6y-72+4y=548
10y=620
y=62

Then there are 62 student tickets and 50 adult tickets.

7 0
3 years ago
Read 2 more answers
The equation would be 10x+20y=150 and if she buys just t- shirts and no jeans she will be able to buy 15 t- shirts
dlinn [17]
10x=150 so this is the answer it should be because this was the right answer on mine
5 0
3 years ago
Math 100 points ill give brainliest as well
padilas [110]

i cant really see the picture well, sorry.

3 0
3 years ago
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