111 lines
No EOL
2.7 KiB
Markdown
111 lines
No EOL
2.7 KiB
Markdown
# Exercises
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So this lesson only had 4 exercises (from what I could see, anyway), but for some reason they start at 16.
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At any rate, this lesson was about modules. `math` and `random` were at the focus of the lesson and its exercises.
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Problem 16
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----------
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_Use a_ `for` _statement to print 10 random numbers._
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### My Code
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```text-x-python
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# Runestone.Academy thinkcspy course
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# Chapter 5
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# Problem 16
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import random
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for i in range(10):
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prob = random.random()
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print(i+1, '-', prob)
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```
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### Result
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Result from the first run of this script
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Problem 17
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----------
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_Repeat the above exercise but this time print 10 random numbers between 25 and 35, inclusive._
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### My Code
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```text-x-python
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# Runestone.Academy thinkcspy course
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# Chapter 5
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# Problem 17
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import random
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for i in range(10):
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prob = random.randrange(25, 36)
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print(i+1, '-', prob)
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```
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### Result
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Result from the first run of this script
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Problem 18
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----------
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_The **Pythagorean Theorem** tells us that the length of the hypotenuse of a right triangle is related to the lengths of the other two sides. Look through the_ `math` _module and see if you can find a function that will compute this relationship for you. Once you find it, write a short program to try it out._
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### My Code
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```text-x-python
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# Runestone.Academy thinkcspy course
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# Chapter 5
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# Problem 18
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import math
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a = float(input('What is the length of the first side of the triangle?'))
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b = float(input('What is the length of the second side of the triangle?'))
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c = math.sqrt(a*a + b*b)
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print('The hypotenuse of this triangle with is', c, '.')
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```
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### Result
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The numbers used to get this result were 12 and 12.
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Problem 19
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----------
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_Search on the internet for a way to calculate an approximation for **pi**. There are many that use simple arithmetic. Write a program to compute the approximation and then print that value as well as the value of_ `math.pi` _from the math module._
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### My Code
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```text-x-python
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# Runestone.Academy thinkcspy course
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# Chapter 5
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# Problem 19
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import math
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c = float(input("Today we are going to calculate pi from the dimensions of a circle. What is the circumference of your circle?"))
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d = float(input("And what is the diameter of your circle?"))
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zpi = c/d
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print("The approximate value of pi from the dimensions provided is", zpi)
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print("The actual (also technically approximate) value of pi is", math.pi)
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```
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### Result
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The numbers were 22.2817 for the diameter and 70 for the circumference. The numbers were generated on [this site](https://www.omnicalculator.com/math/circle-measurements). |