In this Data analysis assignment, we delve into the dynamics of exponential growth and decay through practical scenarios. From the projection of birdhouse production to estimating Christmas party guests and understanding radioactive decay, these examples illustrate the application of mathematical models in various contexts. Let's explore the solutions step by step and witness the power of exponential models in capturing real-world phenomena.

### Charlie's Birdhouses Growth

**Problem Description:** Charlie's Birdhouses experiences an annual growth rate of 10%. In 2010, Charlie made 120 birdhouses. Determine the number of birdhouses he will make in 2015.

**Solution:** The exponential model for the number of birdhouses (Xj) each year is given by

X_(j+1)=1.1×X_j

And X_2010=120. Therefore, X_2015=120×〖1.1〗^5=193.26

Rounding to two decimal places, the expected number of birdhouses in 2015 is approximately 193.

Year | Number of Birdhouses |

2010 | 120.00 |

2011 | 132.00 |

2012 | 145.20 |

2013 | 159.72 |

2014 | 175.69 |

2015 | 193.26 |

**Table 1: Number of Birdhouses vs. the year**

### Karen's Christmas Party Guests

**Problem Description:** Karen's Christmas party started in 1990 with 12 guests, increasing by 5% every other year. Find the expected number of guests for her 2012 Christmas party.

**Solution:** The number of guests (Nj) follows a geometric progression

N_j=1.05×N_(j-1)

The number of guests for 2012 (2012N2012) is found to be 35.1.

Year | No. of Guests |
---|---|

1990 | 12 |

1991 | 12.60 |

1992 | 13.23 |

1993 | 13.89 |

1994 | 14.59 |

1995 | 15.32 |

1996 | 16.08 |

1997 | 16.89 |

1998 | 17.73 |

1999 | 18.62 |

2000 | 19.55 |

2001 | 20.52 |

2002 | 21.55 |

2003 | 22.63 |

2004 | 23.76 |

2005 | 24.95 |

2006 | 26.19 |

2007 | 27.50 |

2008 | 28.88 |

2009 | 30.32 |

2010 | 31.84 |

2011 | 33.43 |

2012 | 35.10 |

**Table 2:** Number of guests vs. the year for party guests

### Uranium-234 Decay

**Problem Description:**Uranium-238, when mined for nuclear reactors, produces Uranium-234 with a half-life of 24 days. Calculate the remaining amount on December 31, 2011, if there were initially 24.5g on January 1, 2011.

**Solution:** Using the radioactive decay formula, the remaining amount (N) is found to be 0.00064g on December 31, 2011.

N = N_0× (1/2) ^(t/T)

N = N0 * (1/2) ^(t/T) = 24.5 * (1/2) ^15.21= 0.00064 g

### Ornithine Decarboxylase Decay

**Problem Description:** The enzyme ornithine decarboxylase in the human body has a half-life of 11 minutes. If 131mg is present, determine the amount remaining after 1 hour.

**Solution:**Applying the radioactive decay formula, approximately 2.98mg of ornithine decarboxylase will remain after 1 hour.

N = N_0× (1/2) ^(t/T)

N = N0 * (1/2) ^(t/T) = 131 * (1/2) ^ (60/11) ≈ 2.98 mg

Therefore, after 1 hour, approximately 2.98 mg of ornithine decarboxylase will be left from the initial 131 mg.

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