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   "source": [
    "# Lab 1\n",
    "(Prof. Carlos J. Costa)"
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**1)**  Verify if  a value is integer"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
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   "cell_type": "markdown",
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   "source": [
    "**2)** Verify if a value is even"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**3)** Insert two numbers. Is the first is bigger than the second?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**4)** Verify if one value is multiple of another"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**5)** Calculate the interest earn by an investor that \n",
    "invested a capital of 200 during 3 years with an interest rate of 3%. \n",
    "(I= P\\*R\\*T)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**6)** Capital that an investor obtained after \n",
    "investing a capital of 200 during 3 years with an interest rate of 3%. (Compound interest)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**7)** Calculate your BMI (Body Mass Index)  \n",
    "   \n",
    "$BMI = mass (kg)/height^2 (m)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**8)** Calcule the Golden ration:   \n",
    "    \n",
    "$gr=(1+\\sqrt{5})/2$\n",
    "\n",
    "1. Solve the problem without using libraries\n",
    "2. Use module math (import math) and function sqr (math.sqrt)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**9)** Calculete the NPV (Net present value) of an investment, considering an initial investment of 10000, the following Cashflows 2000,3000, 4000, 4000 and 5000 and a discount rate of 10%.\n",
    "\n",
    "$NPV = \\sum_{t=1}^{n}\\frac{FV_{t}}{(1 + k)^{t}} – I$\n",
    "\n",
    "Where:\n",
    "\n",
    "FV = Future cost of the cash inflows,\n",
    "I = Initial Investment\n",
    "k = Discount rate equal to the owner’s cost of capital\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**10)** Ask the user to insert name and age. Calculate the the birth. Print a result saying the 'this person was born in'\n",
    "1. Solve the problem without using modules and libraries\n",
    "2. Solve the problem using the date library from module datetime, as fallow:\n",
    "   \n",
    "```python\n",
    "from datetime import date   \n",
    "today = date.today()   \n",
    "today.year   \n",
    "```"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**11)** Ask the user to insert forenames, surnames. create a new variable (name) with your complete name.\n",
    "   \n",
    "Create the follwoing variables:   \n",
    "   \n",
    "**nameBig**, where all the characters of the name are capitalized   \n",
    "**nameTitle**, where the only the first character of each name (word) is capitalized   \n",
    "**nameSmall**, where all the characters of the name are lower   \n",
    "**nameCapitalized**, where only the first charater of the first name is capitalized   "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**12)** Use the follwoing method to show where in which carater appears the firs \"da\"\n",
    "   \n",
    "```python\n",
    "str.find(sub,start,end)\n",
    "```\n",
    "   \n",
    "What happens if does not find?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#Code here"
   ]
  }
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