The annual returns will be calculated as follows:
a] Here we use the formula:
A=p(1+r/100)^n
A=future amount
p=principle
r=returns
n=time
We are given:
A=500, p=400, t=1
Plugging the values in the formula we obtain:
500=400(1+r)^1
simplifying and solving for r:
1.25=1+r
thus
r=1.25-1
r=0.25~25%
b] Using the formula above:
A=p(1+r/100)^n
A=2500+100=2600, p=2000, n=1 year
plugging the values in the equation we obtain:
2600=2000(1+r)^1
simplifying and solving for r we obtain:
2600/2000=1+r
1.3=1+r
hence
r=1.3-1
r=0.3~30%
Answer:
(4x - 3)(3x² + 1)
Step-by-step explanation:
Given
12x³ - 9x² + 4x - 3 ( factor the first/second and third/fourth terms )
= 3x²(4x - 3) + 1(4x - 3) ← factor out (4x - 3) from each term
= (4x - 3)(3x² + 1) ← in factored form
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:
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhlb
Step-by-step explanation:
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