Answer:
- 5 min: 3,029,058
- 10 min: 3,398,220
- 60 min: 10,732,234
Step-by-step explanation:
The given function is evaluated by substituting the given values of t. This requires using the exponential function of your calculator with a base of 'e'. Many calculators have that value built in, or have an e^x function (often associated with the Ln function).
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<h3>5 minutes</h3>
The number of bacteria present after 5 minutes is about ...
f(5) = 2.7×10^6×e^(0.023×5) ≈ 3,029,058
<h3>10 minutes</h3>
The number of bacteria present after 10 minutes is about ...
f(10) = 2.7×10^6×e^(0.023×10) ≈ 3,398,220
<h3>60 minutes</h3>
The number of bacteria present after 60 minutes is about ...
f(60) = 2.7×10^6×e^(0.023×60) ≈ 10,732,234
Answer:
HI! I think the answer is 4.
Step-by-step explanation:
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Step-by-step explanation:
Substitute r = 3 and h = 5 into the formula. Volume V=13⋅π⋅9⋅5=15πcm3=47.1cm3.
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