Soulver33.0.0TNT.zip (12.84 MB) Choose free or premium download SLOW DOWNLOAD. FAST INSTANT DOWNLOAD Download type. Soulver 3.0.1 Multilingual macOS 13 mb. Soulver 3 is a smart notepad with a built-in calculator. It gives instant answers to any calculations it finds in your text. It's a better way to work stuff out than with a traditional calculator, and a more lightweight tool for quick calculations than a spreadsheet.
3Ds Portal » Software » Software MAC » Soulver 3.0.1 Multilingual macOS. Soulver 3.0.1 Multilingual macOS. Soulver 3.0.1 Multilingual macOS 13 mb. Soulver 3 is a smart notepad with a built-in calculator. It gives instant answers to any calculations it finds in your text. It's a better way to work stuff out than with a.
Soulver 3 is a smart bloc notes with a built -in calculator. It gives instant answers to any calculation you find in your text. It’s a better way to solve things with a traditional calculator and a lighter tool faster than a spreadsheet calculations.
Math features
• instant calculations, are not required equal
Use words • next to the numbers, so that the calculations make sense
• fast all your lines Totals and subtotals.
• calculations calendar (June 9 + 3 weeks 3:35 pm + 6 hours 27 minutes)
• easy Ratios ( “$ 300-10%”, “30% as 200”)
• Unit Conversions ( “10 kg in pounds, “25 meters feet”)
• 168 cryptocurrency rates in real time and live.
• Line references to build small spreadsheet documents.
• Variables and autocomplete variables.
• Global variables and custom units
Application Features
• obscurely
• Sidebar to organize your sheets and the ability to make folders
• Headers (#), comments (//) and labels (:)
• Customize your syntax coloring and font
• automatic formatting of numbers, with spaces around operators and thousands separators.
• Scrubbing of numbers (hold down the Shift key and scroll over a number)
• Deep integration of macOS system: compatibility with the touchbar, Quicklook, Calculate services anywhere
• Automation tools: A command-line interface, Alfred workflow, Automator action
'Bi' means 'two' (like a bicycle has two wheels) .. |
Tossing a Coin:
We say the probability of the coin landing H is ½
And the probability of the coin landing T is ½
Throwing a Die:
We say the probability of a four is 1/6 (one of the six faces is a four)
And the probability of not four is 5/6 (five of the six faces are not a four)
Note that a die has 6 sides but here we look at only two cases: 'four: yes' or 'four: no'
Toss a fair coin three times .. what is the chance of getting two Heads?
Tossing a coin three times (H is for heads, T for Tails) can get any of these 8 outcomes:
HHH |
HHT |
HTH |
HTT |
THH |
THT |
TTH |
TTT |
'Two Heads' could be in any order: 'HHT', 'THH' and 'HTH' all have two Heads (and one Tail).
So 3 of the outcomes produce 'Two Heads'.
Each outcome is equally likely, and there are 8 of them, so each outcome has a probability of 1/8
So the probability of event 'Two Heads' is:
Number of outcomes we want | Probability of each outcome | ||
3 | × | 1/8 | = 3/8 |
So the chance of getting Two Heads is 3/8
We used special words:
The calculations are (P means 'Probability of'):
We can write this in terms of a Random Variable, X, = 'The number of Heads from 3 tosses of a coin':
And this is what it looks like as a graph:
It is symmetrical!
Now imagine we want the chances of 5 heads in 9 tosses: to list all 512 outcomes will take a long time!
So let's make a formula.
In our previous example, how can we get the values 1, 3, 3 and 1 ?
Online wing ide. Well, they are actually in Pascal’s Triangle !
Can we make them using a formula?
Sure we can, and here it is:
It is often called 'n choose k'
You can read more about it at Combinations and Permutations.
Let's try it:
We have n=3 and k=2:
So there are 3 outcomes that have '2 Heads'
(We knew that already, but now we have a formula for it.)
Let's use it for a harder question:
We have n=9 and k=5:
So 126 of the outcomes will have 5 heads
And for 9 tosses there are a total of 29 = 512 outcomes, so we get the probability:
Number of outcomes we want | Probability of each outcome | |||
126 | × | 1512 | = | 126512 |
So:
P(X=5) = 126512 = 0.24609375
About a 25% chance.
(Easier than listing them all.)
So far the chances of success or failure have been equally likely.
But what if the coins are biased (land more on one side than another) or choices are not 50/50.
This is just like the heads and tails example, but with 70/30 instead of 50/50.
Let's draw a tree diagram:
The 'Two Chicken' cases are highlighted.
The probabilities for 'two chickens' all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case. In other words
0.147 = 0.7 × 0.7 × 0.3
Or, using exponents:
= 0.72 × 0.31
The 0.7 is the probability of each choice we want, call it p
The 2 is the number of choices we want, call it k
And we have (so far):
= pk × 0.31
The 0.3 is the probability of the opposite choice, so it is: 1−p
The 1 is the number of opposite choices, so it is: n−k
Drawoutx 1 9 1 – scrapbook layout application. Which gives us:
= pk(1-p)(n-k)
Where
So we get:
which is what we got before, but now using a formula
Now we know the probability of each outcome is 0.147
But we need to include that there are three such ways it can happen: (chicken, chicken, other) or (chicken, other, chicken) or (other, chicken, chicken)
The total number of 'two chicken' outcomes is:
And we get:
Number of outcomes we want | Probability of each outcome | |||
3 | × | 0.147 | = | 0.441 |
So the probability of event '2 people out of 3 choose chicken' = 0.441
OK. That was a lot of work for something we knew already, but now we have a formula we can use for harder questions.
So we have:
And we get:
That is the probability of each outcome.
And the total number of those outcomes is:
And we get:
Number of outcomes we want | Probability of each outcome | |||
120 | × | 0.0022235661 | = | 0.266827932 |
So the probability of 7 out of 10 choosing chicken is only about 27%
Moral of the story: even though the long-run average is 70%, don't expect 7 out of the next 10.
Now we know how to calculate how many:
n!k!(n-k)!
And the probability of each:
pk(1-p)(n-k)
When multiplied together we get:
Probability of k out of n ways:
P(k out of n) = n!k!(n-k)!pk(1-p)(n-k)
The General Binomial Probability Formula
Important Notes:
Have a play with the Quincunx (then read Quincunx Explained) to see the Binomial Distribution in action.
A fair die is thrown four times. Calculate the probabilities of getting:
In this case n=4, p = P(Two) = 1/6
X is the Random Variable ‘Number of Twos from four throws’.
Substitute x = 0 to 4 into the formula:
P(k out of n) = n!k!(n-k)! pk(1-p)(n-k)
Like this (to 4 decimal places):
Summary: 'for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)'
This time the graph is not symmetrical:
It is not symmetrical!
It is skewed because p is not 0.5
Your company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed).
What is the expected Mean and Variance of the 4 next inspections?
First, let's calculate all probabilities.
Wimoweh 1 1 68 full. X is the Random Variable 'Number of passes from four inspections'.
Substitute x = 0 to 4 into the formula:
P(k out of n) = n!k!(n-k)! pk(1-p)(n-k)
Like this:
Summary: 'for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection.'
Let's calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.
There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!
The mean, or 'expected value', is:
μ = np
For the sports bikes:
μ = 4 × 0.9 = 3.6
So we can expect 3.6 bikes (out of 4) to pass the inspection.
Makes sense really .. 0.9 chance for each bike times 4 bikes equals 3.6
The formula for Variance is:
Variance: σ2 = np(1-p)
And Standard Deviation is the square root of variance:
σ = √(np(1-p))
For the sports bikes:
Variance: σ2 = 4 × 0.9 × 0.1 = 0.36
Standard Deviation is:
σ = √(0.36) = 0.6
Note: we could also calculate them manually, by making a table like this:
X | P(X) | X × P(X) | X2 × P(X) |
0 | 0.0001 | 0 | 0 |
1 | 0.0036 | 0.0036 | 0.0036 |
2 | 0.0486 | 0.0972 | 0.1944 |
3 | 0.2916 | 0.8748 | 2.6244 |
4 | 0.6561 | 2.6244 | 10.4976 |
SUM: | 3.6 | 13.32 |
The mean is the Sum of (X × P(X)):
μ = 3.6
The variance is the Sum of (X2 × P(X)) minus Mean2:
Variance: σ2 = 13.32 − 3.62 = 0.36
Standard Deviation is:σ = √(0.36) = 0.6
And we got the same results as before (yay!)P(k out of n) = n!k!(n-k)! pk(1-p)(n-k)