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ValentinkaMS [17]
3 years ago
9

The perimeter of a square is 36 inches. Find the length of a diagonal

Mathematics
1 answer:
sergey [27]3 years ago
4 0
The length of a diagonal is nine inches because it is the same length as the side length.
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Please help, picture included!
Mariana [72]
Check the picture, notice the domain in red.

recall that the domain is the values for the x-coordinate, whilst the range is the values for the y-coordinate.

4 0
3 years ago
K to the power of 2-2k-24
insens350 [35]

Answer:

I don’t understand what you are asking

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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
The complement of 20°17' is
mel-nik [20]

Answer:69°43'

Step-by-step explanation:

Complementary angles add up to 90

Let them complement be y

y+20°17`=90°

Collect like terms

y=90-20°17' 20°17'=1217/60

y=90-1217/60

y=(60x90 -1 x 1217)/60

y=(5400-1217)/60

y=4183/60

y=69°43'

7 0
3 years ago
A random sampling of a company's monthly operating expenses for n=36 months produced a sample mean of $5474 and a stan-dard devi
lilavasa [31]

Answer with explanation:

Mean of the sample(m) = $ 5474

Standard deviation of the sample (S)=764

Number of observation(n)=36

Z_{90 \text{Percent}}=Z_{0.09}=0.5359

z_{score}=\frac{\Bar x-\mu}{\frac{S}{\sqrt{n}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{\sqrt{36}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{6}}\\\\764 \times 0.5359=6 \times (5474- \mu)\\\\409.4276=32844-6 \mu\\\\6 \mu=32844 -409.4276\\\\ 6 \mu=32434.5724\\\\ \mu=\frac{32434.5724}{6}\\\\ \mu=5405.76

So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.

4 0
3 years ago
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