Question

PLEASE HELP ASAP!!!!

kathy collected pokemon Go creatures over the course of 5 days. Below is a chart that displayed the number of creatures she collected each day
Day 1 Day2. Day3. Day4. Day5.
42. 26. 61. 65. 56.

A. How many creatures would you expect Kathy to collect on day 6? Explain your prediction?

B. If Kathy collected 53 Pokémon creatures, how would her average change?

C. What are the number of creatures she collected on day 6, 7, and 8?

Answer 1

Let us model this problem with a polynomial function.

Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.

Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅

Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |

When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000

Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56

The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.

Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631

Answer 2

Given: Data for 5 Days

Part A:-

No. of creatures on 6th day = Average/Mean of the Given Data

`\text{Mean = }\frac{\text{Sum of all the observation}}{\text{Total No. of observation}}`

⇒
`\text{Mean = }(42+26+61+65+56)/(5)`

⇒ Mean = 50

∴She collect 50 creatures on day 6.

Part B:-

As we Predicted she collected 50 pokenmon on day 6. So her Average is 50.

if She collected 53 creatures, then the average will be-

`\text{Average = }(42+26+61+65+56+53)/(6)`

⇒ Average = 50.5

Therefore, her new average become 50.5

Part C:-

There is no fixed no of creature collected on day 6,7,8. It can be increased or Decreased but her average will always be around 50.