Answer:
coefficient of the term of degree 4 in this polynomial is 3
Step-by-step explanation:
Answer:
One
General Formulas and Concepts:
<u>Algebra I</u>
- Reading a coordinate plane
- Solving systems of equations by graphing
Step-by-step explanation:
If 2 lines are parallel, they will have no solution.
If 2 lines are the same, they will have infinite amount of solutions.
We see from the graph that the 2 lines intersect at one point, near x = 1.5.
∴ our systems has 1 solution.
Well, there are 60 minutes in 1 hour, and there are 60 seconds in 1 minute, so in 60 minutes, there are 60 * 60 seconds, or 3600 seconds in 60 minutes, or 1 hour, therefore, in 1 hour there are 3600 seconds.
the machine makes 7 aliens in 2 seconds, how many will it make in 3600 seconds?
![\bf \begin{array}{ccll} aliens&seconds\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 7&2\\ a&3600 \end{array}\implies \cfrac{7}{a}=\cfrac{2}{3600}\implies \cfrac{7\cdot 3600}{2}=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%0Aaliens%26seconds%5C%5C%0A%5Ctext%7B%5Ctextemdash%5Ctextemdash%5Ctextemdash%7D%26%5Ctext%7B%5Ctextemdash%5Ctextemdash%5Ctextemdash%7D%5C%5C%0A7%262%5C%5C%0Aa%263600%0A%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B7%7D%7Ba%7D%3D%5Ccfrac%7B2%7D%7B3600%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%203600%7D%7B2%7D%3Da)
that many.
Answer:
Step-by-step explanation:
Suppose we think of an alphabet X to be the Event of the evidence.
Also, if Y be the Event of cheating; &
Y' be the Event of not involved in cheating
From the given information:
![P(\dfrac{X}{Y}) = 60\% = 0.6](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BX%7D%7BY%7D%29%20%3D%2060%5C%25%20%3D%200.6)
![P(\dfrac{X}{Y'}) = 0.01\% = 0.0001](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BX%7D%7BY%27%7D%29%20%3D%200.01%5C%25%20%3D%200.0001)
![P(Y) = 0.01](https://tex.z-dn.net/?f=P%28Y%29%20%3D%200.01)
Thus, ![P(Y') \ will\ be = 1 - P(Y)](https://tex.z-dn.net/?f=P%28Y%27%29%20%5C%20will%5C%20be%20%3D%201%20-%20P%28Y%29)
P(Y') = 1 - 0.01
P(Y') = 0.99
The probability of cheating & the evidence is present is = P(YX)
![P(YX) = P(\dfrac{X}{Y}) \ P(Y)](https://tex.z-dn.net/?f=P%28YX%29%20%3D%20P%28%5Cdfrac%7BX%7D%7BY%7D%29%20%5C%20P%28Y%29)
![P(YX) =0.6 \times 0.01](https://tex.z-dn.net/?f=P%28YX%29%20%3D0.6%20%5Ctimes%200.01)
![P(YX) =0.006](https://tex.z-dn.net/?f=P%28YX%29%20%3D0.006)
The probabilities of not involved in cheating & the evidence are present is:
![P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})](https://tex.z-dn.net/?f=P%28Y%27X%29%20%3D%20P%28Y%27%29%20%20%5Ctimes%20P%28%5Cdfrac%7BX%7D%7BY%27%7D%29)
![P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099](https://tex.z-dn.net/?f=P%28Y%27X%29%20%3D%200.99%20%20%5Ctimes%200.0001%20%5C%5C%20%5C%5C%20%20P%28Y%27X%29%20%3D%200.000099)
(b)
The required probability that the evidence is present is:
P(YX or Y'X) = 0.006 + 0.000099
P(YX or Y'X) = 0.006099
(c)
The required probability that (S) cheat provided the evidence being present is:
Using Bayes Theorem
![P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BY%7D%7BX%7D%29%20%3D%20%5Cdfrac%7BP%28YX%29%7D%7BP%28Y%29%7D)
![P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BY%7D%7BX%7D%29%20%3D%20%5Cdfrac%7BP%280.006%29%7D%7BP%280.006099%29%7D)
![P(\dfrac{Y}{X}) = 0.9838](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BY%7D%7BX%7D%29%20%3D%200.9838)