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mathisfun.com/geometry/construct-linebisect.html
Here is a useful link to the correct steps. The instructions may not be exactly the same but I think you can do it.
The probability that a 2 or 3 is rolled at least once is 0.14.
<h3>
Probability of not rolling a 2 or 3 in a single die roll of two dice</h3>
when a six die is rolled once there is 11/12 probability of not rolling 2 or 3.
rolling a six sided die two times, p(2 or 3)' = 11/12
rolling a six sided die four times, p(2 or 3)' = (11/12)² = 121/144
the probability that a 2 or 3 is rolled at least once is calculated as follows;
P = 1 - p(2 or 3)'
P = 1 - 121/144
P = 20/144
P = 0.14
Thus, the probability that a 2 or 3 is rolled at least once is 0.14.
Learn more about probability here: brainly.com/question/24756209
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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
F(x)=2x^2-x-6
Factoring:
f(x)=2(2x^2-x-6)/2=(2^2x^2-2x-12)/2=[(2x)^2-(2x)-12]/2
f(x)=(2x-4)(2x+3)/2=(2x/2-4/2)(2x+3)→f(x)=(x-2)(2x+3)
g(x)=x^2-4
Factoring
g(x)=[sqrt(x^2)-sqrt(4)][sqrt(x^2)+sqrt(4)]
g(x)=(x-2)(x+2)
f(x)/g(x)=[(x-2)(2x+3)] / [(x-2)(x+2)
Simplifying:
f(x)/g(x)=(2x+3)/(x+2)
Answer: Third Option (2x+3)/(x+2)
