Remove the IPYNB Checkpoints, Did Week 5 HW
-- IPYNB Checkpoints ++ Git Ignore File for IPYNB Checkpoints +- Week 5 Homework Files
This commit is contained in:
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4fc499169b
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7 changed files with 219 additions and 60 deletions
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@ -20,11 +20,22 @@
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"execution_count": 4,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2.7182820533299275"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"roughly_e = ...\n",
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"roughly_e = (1 + (1/99999999999))**99999999999\n",
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"roughly_e"
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]
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},
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@ -39,19 +50,30 @@
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"execution_count": 6,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2.7182820533299275"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"n = ...\n",
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"also_roughly_e = ...\n",
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"n = 99999999999\n",
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"also_roughly_e = (1+(1/n))**n\n",
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"also_roughly_e"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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@ -65,7 +87,7 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.3"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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@ -40,20 +40,20 @@
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},
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"outputs": [],
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"source": [
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"a = 1410\n",
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"b = 1.989 * 10**30\n",
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"c = 3\n",
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"d = b / a\n",
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"e = 3.14159\n",
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"f = (d * 3/(4*e))**(1/3)\n",
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"g = 4 * e * f**2\n",
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"h = c * g"
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"sun_density = 1410\n",
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"sun_weight = 1.989 * 10**30\n",
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"depth = 3\n",
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"sun_volume = sun_weight / sun_density\n",
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"pi = 3.14159\n",
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"sun_radius = (sun_volume * 3/(4*pi))**(1/3)\n",
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"sun_surface_area = 4 * pi * sun_radius**2\n",
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"required_beer_volume = depth * sun_surface_area"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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@ -67,7 +67,7 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.1"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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@ -44,20 +44,20 @@
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"outputs": [],
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"source": [
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"two = 2\n",
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"four = two plus two"
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"four = two + two"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"*Write your answer here, replacing this text.*"
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"Python didn't recognize the value called 'plus' and was probably confused about it being there, and since 'plus' is a mathmatical operation, I replaced 'plus' with '+' which when ran, gave no errors."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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@ -71,7 +71,7 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.1"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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@ -21,11 +21,22 @@
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"execution_count": 7,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"0.9999961853027344"
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]
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},
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"execution_count": 7,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"..."
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"1 - (2**2 / 2**20)"
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]
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},
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{
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@ -39,21 +50,35 @@
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"cell_type": "code",
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"execution_count": 9,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"29.0"
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]
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},
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"execution_count": 9,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"1 - 2 ** 3 / 4 + 5 * 6"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": true
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},
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"execution_count": 8,
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your prediction here:\n",
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"..."
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"# Your prediction here: 29\n",
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"# Order of Operations: (L-R)\n",
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"# 2 ** 3 = 8\n",
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"# 8 / 4 = 2\n",
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"# 5 * 6 = 30\n",
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"# 1 - 2 = -1\n",
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"# -1 + 30 = 29"
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]
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},
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{
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@ -67,7 +92,18 @@
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"cell_type": "code",
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"execution_count": 10,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-1805811300.4191697"
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]
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},
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"execution_count": 10,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"1 - 2 ** (3 / 4 + 5 * 6)"
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]
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@ -76,7 +112,9 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"*Write your answer here, replacing this text.*"
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"Prediction: The Value will be smaller, since everything in the parentheses will be done first, then that value will be the exponent for 2, aka 2 ^ value_in_parentheses, and then one will be added to that value(or subtracted depending on how you look at it), and will be shown.\n",
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"\n",
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"After cell execution: Right, and that is a rather large negative number."
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]
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},
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{
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@ -94,17 +132,36 @@
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"execution_count": 25,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2.7182216960543433"
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]
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},
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"execution_count": 25,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"..."
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"n = 1000000000000\n",
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"(1-(1/n))**-n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"When you enter in 10000 for n, you get the 'rough' 2.718, and by adding more zeros, you get a value more precise. However, after adding more than 12 zeros, that number seems to make the rounding happen in a way that the value is no longer 2.718, but 2.719 and increasing the more zeros added..."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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@ -118,7 +175,7 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.3"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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@ -24,7 +24,8 @@
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},
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"outputs": [],
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"source": [
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"a + b + c - max(a, b, c) + min(a, b, c)"
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"a + b + c - (max(a, b, c) + min(a, b, c))\n",
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"# Originally: a + b + c - max(a, b, c) + min(a, b, c)"
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]
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},
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{
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@ -38,15 +39,47 @@
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 6,
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"metadata": {},
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"outputs": [],
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"source": []
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"outputs": [
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{
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"data": {
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"text/plain": [
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"888"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Testing the interesting function: a + b + c - max(a, b, c) + min(a, b, c)\n",
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"\n",
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"# Using the test input vals a = 4, b = 2, c = 7\n",
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"\n",
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"##a = 4\n",
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"##b = 2\n",
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"##c = 7\n",
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"\n",
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"a = 999\n",
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"b = 345\n",
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"c = 888\n",
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"\n",
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"a + b + c - (max(a, b, c) + min(a, b, c))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Code was not correct, would return a value that is not the middle one. By adding parentheses before the call of max and after the call of min, you can return a middle value of three variables every single time."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.3"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"1.0"
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]
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},
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"execution_count": 1,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"rounded = ...\n",
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"# Round can take two parameters, the first is the number to round, while the second is the decimal places in which to round to.\n",
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"rounded = round(1.23456789, 0)\n",
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"rounded"
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]
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},
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"1.235"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"less_rounded = ...\n",
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"# Using round, with the second parameter as three, you get three decimal point precision.\n",
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"less_rounded = round(1.23456789, 3)\n",
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"less_rounded"
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]
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},
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"71.0"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sum_of_rounded_amounts = ...\n",
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"sum_of_rounded_amounts = round(32.25, 0) + round(15.00, 0) + round(24.45, 0)\n",
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"sum_of_rounded_amounts"
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]
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},
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},
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{
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"cell_type": "code",
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"execution_count": 9,
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"execution_count": 4,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"72.0"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"rounded_sum = ...\n",
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"rounded_sum = round((32.25 + 15.00 + 24.45), 0)\n",
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"rounded_sum"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.3"
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"version": "3.9.12"
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}
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},
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"nbformat": 4,
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