Math Problem Statement

Input values into Z ranging from 100-300 in multiples of 20 then solve for p/z. Afterwards, show the percentage increase between the previous values of p/z.

p = ((.5*z)-9)*122.5)) then add 2025 to the final value to get P

Solution

Let's solve the problem with the corrected formula for pp.

Given:

  • z=Bullets firedz = \text{Bullets fired} (values from 100 to 300 in multiples of 20)

The formula for total damage pp is: p=((0.5z9)×122.5)+2025p = \left( (0.5 \cdot z - 9) \times 122.5 \right) + 2025

We will calculate pz\frac{p}{z} for zz ranging from 100 to 300 in multiples of 20, and then calculate the percentage increase between successive values.

Calculations

For z=100z = 100:

p=((0.51009)×122.5)+2025p = \left( (0.5 \cdot 100 - 9) \times 122.5 \right) + 2025 p=((509)×122.5)+2025p = \left( (50 - 9) \times 122.5 \right) + 2025 p=(41×122.5)+2025p = (41 \times 122.5) + 2025 p=5022.5+2025p = 5022.5 + 2025 p=7047.5p = 7047.5 pz=7047.5100=70.475\frac{p}{z} = \frac{7047.5}{100} = 70.475

For z=120z = 120:

p=((0.51209)×122.5)+2025p = \left( (0.5 \cdot 120 - 9) \times 122.5 \right) + 2025 p=((609)×122.5)+2025p = \left( (60 - 9) \times 122.5 \right) + 2025 p=(51×122.5)+2025p = (51 \times 122.5) + 2025 p=6247.5+2025p = 6247.5 + 2025 p=8272.5p = 8272.5 pz=8272.512068.9375\frac{p}{z} = \frac{8272.5}{120} \approx 68.9375

For z=140z = 140:

p=((0.51409)×122.5)+2025p = \left( (0.5 \cdot 140 - 9) \times 122.5 \right) + 2025 p=((709)×122.5)+2025p = \left( (70 - 9) \times 122.5 \right) + 2025 p=(61×122.5)+2025p = (61 \times 122.5) + 2025 p=7472.5+2025p = 7472.5 + 2025 p=9497.5p = 9497.5 pz=9497.514067.8375\frac{p}{z} = \frac{9497.5}{140} \approx 67.8375

For z=160z = 160:

p=((0.51609)×122.5)+2025p = \left( (0.5 \cdot 160 - 9) \times 122.5 \right) + 2025 p=((809)×122.5)+2025p = \left( (80 - 9) \times 122.5 \right) + 2025 p=(71×122.5)+2025p = (71 \times 122.5) + 2025 p=8697.5+2025p = 8697.5 + 2025 p=10722.5p = 10722.5 pz=10722.516067.015625\frac{p}{z} = \frac{10722.5}{160} \approx 67.015625

For z=180z = 180:

p=((0.51809)×122.5)+2025p = \left( (0.5 \cdot 180 - 9) \times 122.5 \right) + 2025 p=((909)×122.5)+2025p = \left( (90 - 9) \times 122.5 \right) + 2025 p=(81×122.5)+2025p = (81 \times 122.5) + 2025 p=9922.5+2025p = 9922.5 + 2025 p=11947.5p = 11947.5 pz=11947.518066.375\frac{p}{z} = \frac{11947.5}{180} \approx 66.375

For z=200z = 200:

p=((0.52009)×122.5)+2025p = \left( (0.5 \cdot 200 - 9) \times 122.5 \right) + 2025 p=((1009)×122.5)+2025p = \left( (100 - 9) \times 122.5 \right) + 2025 p=(91×122.5)+2025p = (91 \times 122.5) + 2025 p=11147.5+2025p = 11147.5 + 2025 p=13172.5p = 13172.5 [ \frac{p}{z} = \frac{13172.5}{200

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Percentage Increase

Formulas

Linear equation formula: p = (0.5 * z - 9) * 122.5 + 2025
Percentage increase formula: ((new value - old value) / old value) * 100

Theorems

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Suitable Grade Level

Advanced High School