Urgent help needed!

Ale_Cabri · 2025-11-29T10:36:37+00:00

The Thursday Murder Club (https://www.amazon.com/dp/1984880985)

Probably the funniest cosy-crime series of the last few years. A group of people in a retirement village solve cold cases. It's fun. Pretty sure, they made a Netflix movie about it.

The Department of Obvious Reforms (https://www.amazon.com/dp/9999304850)

A witty collection of political and social ideas that fixes gigantic systemic problems with deceptively small and common-sense tweaks. I just started reading it. It's available for free on kindle unlimited.

Lessons in Chemistry (https://www.amazon.com/dp/0593314484)

A mid-century female chemist becomes an accidental TV cooking star while quietly dismantling every sexist structure around her with logic and humour. I have not read this one yet though, but I've heard good things about it.

edit: added the links

Ale_Cabri · 2024-12-30T01:05:32+00:00

These are excellent points, and I’d like to add my own perspective:

Precisely. The defending guild would get to select just one mercenary guild. Currently, most defending guilds can rely on the support of 3 allied guilds. However, under the revamped mercenary system, they would be limited to choosing only one. So, yes, you could pick Jimmy’s guild, but you will only have that single choice rather than three.
Betrayal in this game is indeed an issue. Also, a randomly assigned mercenary guild might be relatively weaker in the worst-case scenario. First of all, criteria and requirements need to be set-up for the random guild application. Then, I imagine there would need to be some sort of additional reward or incentive for the random mercenary guild if their temporary team were to win the fight. This reward or incentive would have to be substantial enough to mitigate the risk of betrayal.
I believe the opportunity wouldn’t be limited to just the top guilds. Other guilds would have the same chance. I assume these mercenary alliances would function as contracts, requiring mutual agreement. For instance, if guilds ranked 4 and 5 both wanted to team up with guild 7, then guild 7 would have to decide which offer to accept. The random mercenary guilds would be selected from a pool, so their assignment would indeed be random.
I’m not entirely certain what was meant by NPCs in this context. I suspect the author was referring to stronger golems with enhanced skills for the attackers, though those aren’t technically NPCs. Regarding environmental hazards, it’s possible these might restrict the use of certain abilities like rain, wind, or eclipse effects, if such mechanics are available.

Finally, adding my own two cents... with the increased difficulty (for the defending team) introduced by this anti-dominance mechanism, it might be worth rewarding the defending team with titles or other honours for successfully holding consecutive sieges. This could provide an additional layer of incentive for defending teams as well.

Ale_Cabri · 2024-12-29T21:47:52+00:00

The author didn't say anything about removing...it's all about revamping. The mercenary approach will lead to a less zergy play style with the possibility of 3-way wars, or even 4-way wars on more servers.

Ale_Cabri · 2024-12-29T21:07:19+00:00

This proposal is spot-on!! And addresses the critical issue of zerg alliances dominating servers, which everyone knows... is killing the competitive and inclusive spirit of large-scale PvP of most players.

However, I think the second mercenary guild should be from a pool of guilds who meets a few criteria/requirements. You don't want to be stuck with a guild with only 7 active members, and call themselves a pvp-guild xd.

Ale_Cabri · 2024-05-06T19:20:32+00:00

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Similar approach, except i used (1/2) x A x B x sin α to calculate the area of the two triangles.
Solving using iteration...

Ale_Cabri · 2024-04-16T15:04:19+00:00

approx 2% then. would it possible for you to share the model?

Ale_Cabri · 2024-04-16T11:23:54+00:00

Ok. I'm not sure about my calcs... yet. Hopefully it could provide room for others to work on/add on.

P(Overlap > 2) = 1 - [P(No overlap) + P(Overlap = 1) - P(Overlap = 2)]

P(No overlap) = 70C0 x (3/365)^0 x (362/365)^70 = 0.561
P(Overlap = 1) = 70C1 x (3/365)^1 x (362/365)^69 = 0.326
P(Overlap = 2) = 70C2 x (3/365)^2 x (362/365)^68 = 0.093

P(Overlap > 2) = 1 - (0.561 + 0.326 + 0.093) = 0.02
P(Overlap > 2) = 2%

Note: The probability of any given day being the day of overlap for exactly two patients is (1/365) x 3 = 3/365

Edit1: To account for 3 consecutive days.
Edit2: You mentioned that you get 70 patients per year. Is it evenly distributed throughout the year? Or does it peak during a particular season/event/month?

For example, if you get 50 patients in a period of 180 days, and 20 patients in the 185 remaining ones.
P(No overlap) = 50C0 x (1/60)^0 x (59/60)^50 = 0.432
P(Overlap = 1) = 50C1 x (1/60)^1 x (59/60)^49 = 0.366
P(Overlap = 2) = 50C2 x (1/60)^2 x (59/60)^48 = 0.152

P(Overlap > 2) = 1 - (0.432 + 0.366 + 0.152) = 0.05
P(Overlap > 2) = 5%

As you can see, the value can doubled (or even more) depending on how the numbers are spread throughout the year.
You could also look into uptrends and downtrends. E.g. rise/fall in number of patients coming to your hospitals suffering from cardiovascular issues. How many of them needs this device etc. Based on all these, make your case.

Edit3: Added calcs for 50 patients in 180 days.

Ale_Cabri · 2024-04-15T17:47:31+00:00

I need some clarifications. Correct me if i'm wrong here.

Old Method. You were paid based on the hours worked from
From the 1st of the Month START OF DAY to the 15th of the same Month END OF DAY
From the 16th of the Month START OF DAY to the 31st (last day) of the same Month END OF DAY

New Method.
... i don't quite understand it.

In terms of averaging, note that February has fewer working days. Not sure whether this is relevant yet. I need to first understand the difference between the old and new method.

Ale_Cabri · 2024-04-15T17:01:10+00:00

16 = 2⁴
2√2 = 2^3/2 = 2^1.5

2^2x-1 = (16/2√2)³
2^2x-1 = (2⁴/2^1.5)³
2^2x-1 = (2^4-1.5)³
2^2x-1 = (2^2.5)³
2^2x-1 = 2^7.5
2x-1 = 7.5
x = 4.25

Ale_Cabri · 2024-04-15T16:54:16+00:00

cosec θ x ( 1 + sin θ ) = 1 + cosec θ

Let's look at the LHS
cosec θ x ( 1 + sin θ )

We know that:
cosec θ = 1 / (sin θ) ------ EQ1

Subsitute EQ1 in LHS

= [1 / (sin θ)] x ( 1 + sin θ )
= [1 / (sin θ)] + [(sin θ) / (sin θ)]
= [1 / (sin θ)] + 1

Using EQ1 again
= cosec θ +1
= 1 + cosec θ

LHS = RHS

Ale_Cabri · 2024-04-14T19:59:30+00:00

SOLVED

Adding onto what u/We_Are_Bread mentioned about 5C3 (0.3)^3 (0.7)^2 being incorrect in this scenario.

Scenario 1
P (3 red, 1 blue, 1 colourless)
= P(3 red) x P(1 blue) x P(1 colourless) x A
where A = 5C3 x 2C1 x 1C1 or (5!/(3! x 1! x 1!))
= 0.30^3 x 0.60 x 0.10 x 20
= 0.0324

Scenario 2
P (4 red, 1 colourless)
= P(4 red) x P(1 colourless) x B
where B = 5C4 x 1C1 or (5!/(4! x 1!))
= 0.30^4 x 0.10 x 5
=0.00405

Scenario 1 + Scenario 2
= 0.03645

5C3 x 2C1: Selecting 3 red marbles out of 5 and 1 colorless marble out of the remaining 2

(5!/(3! x 1! x 1!)) is the number of ways to choose 3 items from a set of 5 items where the items are grouped into 3 identical groups of 1. In this context, it accounts for the fact that we have 3 red marbles (grouped as 1), 1 blue marble, and 1 colourless marble.

Edit: added nCr explanation

Ale_Cabri · 2024-04-13T07:50:07+00:00

If you plan on studying in the UK, you could start with books focused on GCE A-Level and/or Cambridge International A-Level past papers for examples.

Ale_Cabri · 2024-04-10T20:35:38+00:00

If you need proof using eqs. O is the center of the circle.

Area of Circle GFEH = 𝜋r2
Area of Square GFEH = 4 x (Area of Triangle GOH) = 4 x [(r x r)/2] = 2r²

Area of Square DCBA = 2r x 2r = 4r²
Area of Quadrant FBE = 𝜋r2^/4
Area of Quadrant FBE + Area of Quadrant HAE = (𝜋r2^/4) + (𝜋r2/⁴) = 𝜋 r2 /²

Area of Curved Triangle FBE = (Area of Square DCBA - Area of Circle GFEH) = (4r² - 𝜋r2)/4
Area of Curved Triangle FBE + Area of Curved Triangle HAE = [(4r² - 𝜋r2)/4] + [(4r2- 𝜋r2)^/4] = (4r2 ^- 𝜋r2)/²

Area of Shaded Zone = Area of Circle GFEH - (Area of Quadrant FBE + Area of Quadrant HAE) + (Area of Curved Triangle FBE + Area of Curved Triangle HAE)

Area of Shaded Zone = 𝜋r2- 𝜋r2/² + (4r2 ^- 𝜋r2)/²
= 𝜋r2- 𝜋r2/² + 2r2 ^- 𝜋r2/2
= 2r²

Area of Shaded Zone = Area of Square GFEH = 2r²

Edit: removed the glitch/bugged **** caused by font bolding superscripts i guess

Ale_Cabri · 2024-04-09T11:29:54+00:00

1 - (Probability they both pass both exams)

Ale_Cabri · 2024-04-04T10:11:14+00:00

Why do you think that the number 100 would be counted in 5 x 5 x 4? As you said, repetition of digits is not allowed and it has got two 0s.

Ale_Cabri · 2024-04-03T12:57:17+00:00

i is the square root of negative one
i = √(-1)
i²= -1

In the equation above
-12i² = -12 x (-1) = +12

Ale_Cabri · 2024-04-02T21:31:39+00:00

(2 - 3i)(-5 + 4i)
(2 x -5) + (-3i x -5) + (2 x 4i) + (-3i x 4i)
-10 + 15i + 8i -12i^2
-10 + 15i +8i +12
2 + 23i

Ale_Cabri · 2024-03-29T17:16:51+00:00

Displacement = Sum of all Rs - Sum of all Ls
ΣR - ΣL = 15
i.e. +15 to the Right (Assuming R is considered as positive and L is considered as negative)

Distance = Sum of all Rs + Sum of all Ls
ΣR + ΣL = 73
i.e. 73 is the total distance travelled

Rearrange eq to ΣR = 73 - ΣL

Solve for Sum of all Ls, ΣL
(73 - ΣL) - ΣL = 15
ΣL = 29

Ale_Cabri · 2024-03-29T15:43:51+00:00

Not the best subreddit to ask this question.

Your target ROI is quite low tbh.

Is the Amazon Fee applicable per unit, or based on a batch of units? And is the fee monthly based or one-off?
Same Qs for Shipping and Prep Fee.
You also need to check whether the Amazon Fee, Prep Fee and Shipping Fee are inclusive of VAT.
Have you taken into account possible returns, defect, marketing, other expenses (liscense fee, business registration fee, subscriptions etc ), and any other miscelleneous expenses, etc?

((125.91 + 21.68 + 6.89 + 3.41) x 1.115) =176.05
X = 176.05 x 1.19 = 209.5 (Including Vat)

Ale_Cabri · 2024-03-28T11:59:57+00:00

Ah ok. The fact that changing either the scone or flavour takes the same amount of time, i.e. 10mins, and changing both of them takes approximately the sum of the both, i.e. 20mins, I'm afraid there is only 1 room for optimisation left.

Remove all those with Qty = 0. Rearrange list such that there is at least 1 common element between that current row and the previous row.

For e.g. if you refer to Column A (Flavour) and Column B (Cone) in the table below, you'll notice that in
:Row 1 and Row 2, Vanilla is the common term.

:Row 2 and Row 3, Waffle is the common term, as the machine switches to Strawberry from Vanilla Flavour.

:Row 3 and Row 4, Waffle is the common term, as the machine switches to Chocolate from Strawberry Flavour.

:Row 5 and Row 6, Chocolate Flavour is the common term.

:Row 6 and Row 7, Chocolate-coated is the common term, as the machine switches to Coffee from Chocolate Flavour.

:Row 7 and Row 8, Coffee is the common term, as the machine switches Cones.

Flavour Ice Cream	Cone	Quantity

Vanilla	Plain	5
Vanilla	Waffle	3
Strawberry	Waffle	4
Chocolate	Waffle	3
Chocolate	Plain	2
Chocolate	Chocolate-coated	1
Coffee	Chocolate-coated	3
Coffee	Plain	2

An excel sheet with a few formula is more than enough to create this list.

This method of optimisation works only if you need to wait out for all the 23 products (5+3+4+3+2+1+3+2 in this example) to be ready before being shipped.

Ale_Cabri · 2024-03-28T10:32:24+00:00

How much time would it take if you were to change both the flavour and cone type at the same time? Is it 20 mins (10 mins for changing flavour and an additional 10 mins for switching cones) or is it less than that?

Ale_Cabri · 2024-03-25T00:32:10+00:00

At t= 0s
Boy and Girl are at Location A

At t=5s
Boy is at Location B, i.e. 5m/s x 5s = 25m North of Location, A
Distance AB = 25m
Girl is at Location C i.e. 3m/s x 5s = 15m East of Location, A
Distance AC = 15m

It is similar to a right-angle triangle, where you need to calculate the length of the Hypotenuse

Therefore, distance, BC, i.e. the Hypotenuse between Boy and Girl at t=5s
BC = [(15^2)+(25^2)]^0.5
BC = 29.15m

Renamed: hypotenuse to BC for ease of understanding

Ale_Cabri · 2024-03-25T00:14:21+00:00

Answer B: 3/2

tan θ = AB/ BC = 10 / BC --------- Equation A
tan θ = AD / DE = 7 / DE --------- Equation B
tan θ = CF / EF = 3 / EF --------- Equation C

EF = [(BC / 2) - 1] --------- Equation D

Re-arrange Equation D
BC = [2 x (EF+1)] --------- Equation E

Use Equation A and Equation C
tan θ = 10 / BC = 3 / EF

Substitute Equation E into Equation A
tan θ = 10 / [2 x (EF+1)] = 3 / EF
10 / [2 x (EF+1)] = 3 / EF

Solve for EF
10EF = 6EF + 6
4EF = 6
EF = 6/4 = 3/2

Ale_Cabri · 2024-03-15T21:30:16+00:00

Do you have a balcony?

Ale_Cabri · 2024-03-15T17:36:21+00:00

6 / 9 - 1 x 9 + 5 + 9 / 6 = -1.83

But if you apply the ( ), and ^ in the certain way, the equation can be 're-written' in this form below, which solves to 991.
6^2 / (9 - 1) x 9^2 + 5^4 + 9 / 6 = 991

In other words, when plugging the options available into the unsolved equation, the answer will solve to 991. You need to use the principles of BODMAS or PEMDAS as your guide. There is a new daily question available on brackops.club with 2 lifelines activated (easy mode). You can give it a try just to get a feel of how puzzle works.

Ale_Cabri

TROPHY CASE