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Reminders… Hug your folks. Clean up your room.
Test Announcements…
Statistics –
Geometry –
Pre-Calculus –
Extra…
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If you need a calculator, here is the link to a Virtual TI-83 Plus Calculator that I showed you in class.
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Lesson Plans – November 9 – 13, 2009
Pre-Calculus
Standards: Models for Real World Phenomena; Algebraic Functions; Trigonometric Functions; Sequences and Series
Monday, November 9: 4.1 Determine the inverse of a function; Obtain the graph of the inverse function from the graph of the function; Find the inverse function f-1; Review Chapter 3 Part 2
Tuesday, November 10: TEST Chapter 3 Part 2
Wednesday, November 11: 4.2 Evaluate exponential functions; Graph exponential functions; Define the number e; Solve exponential equations
Thursday, November 12: 4.2 Evaluate exponential functions; Graph exponential functions; Define the number e; Solve exponential equations
Friday, November 13: 4.2 Evaluate exponential functions; Graph exponential functions; Define the number e; Solve exponential equations
Geometry
Standards: Number and Operations; Algebra; Geometry; Measurement; Data Analysis and Probability
Monday, November 9: 5.5 Use inequalities involving angles of triangles; Use inequalities involving sides of triangles
Tuesday, November 10: Review Chapter 5
Wednesday, November 11: TEST Chapter 5
Thursday, November 12: 6.2 Parallelograms; Use relationships among sides and among angles of parallelograms; Use relationships involving diagonals of parallelograms or transversals
Friday, November 13: 6.2 Parallelograms; Use relationships among sides and among angles of parallelograms; Use relationships involving diagonals of parallelograms or transversals
Statistics
Standards: Experimental Design; Data Analysis
Monday, November 9: 4.2 Determine if a probability experiment is a binomial experiment; Find binomial probabilities using the binomial probability formula; Construct a binomial distribution and its graph; Find the mean, variance, and standard deviation of a binomial probability experiment
Tuesday, November 10: 4.2 Determine if a probability experiment is a binomial experiment; Find binomial probabilities using the binomial probability formula; Construct a binomial distribution and its graph; Find the mean, variance, and standard deviation of a binomial probability experiment
Wednesday, November 11: 4.3 Find probabilities using the geometric distribution; Find probabilities using the Poisson distribution
Thursday, November 12: 4.3 Find probabilities using the geometric distribution; Find probabilities using the Poisson distribution
Friday, November 13: 4.3 Find probabilities using the geometric distribution; Find probabilities using the Poisson distribution
For all classes.
Activities: Lecture, Board work, Classroom participation
Material: Book problems and Teacher made handouts
Assessment: Daily assignments graded for accuracy. Unit test planned.
If you need a calculator, here is the link to the Virtual TI-83 Plus Calculator that I showed you in class.
The results of a recent survey indicate that 58% of households in the United States own a gas grill. If you randomly select 100 households, what is the probability that exactly 65 households will own a gas grill?
P(65) = 100C65 (.58)65 (.42)35 = 0.0299216472
Let me show you how to use that nice calculator of yours.
Press 2ND DISTR -> :binompdf( -> ENTER
Enter the values for n, p, and r separated by commas
binompdf(100,.58,65) -> ENTER
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What is the probability that at least 65 households will own a gas grill?
P(at least 65) = P(65) + P(66) + P(67) + P(68) + P(69) + P(70) + P(71) +… + P(100)
Press 2ND DISTR -> :binomcdf( -> ENTER
binomcdf(100,.58,64) -> ENTER
CDF means cumulative density function and it allows for the computation for the probability of “r and fewer” successes. It gives you a cumulative total of probabilities. The reason I entered 64 instead of 65 is I wanted to find the probability of 0 successes through 64 successes. I want to subtract the probabilities from 0 through 64. This is the results that is on the screen – this number represents the total cumulative probabilities from 0 to 64.
Now in order to find the probability of at least 65, which is the complement of the event 64 and fewer, I will subtract this result from 1 and the solution is given.
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Example: The results of a recent survey indicate that 71% of people in the United States use more than one topping on their hotdogs. If you randomly select 250 people in the United States, what is the probability that exactly 178 of them will use more than one topping?
binompdf(250,.71,178) = 0.0555119581
What is the probability that at least 178 of them will use more than one topping?
1 – binomcdf(250,.71,177) = 0.5039008221
Geometry
Chapter 5
1. Handout 5310
2. Handout 5210
3. Handout 5110
4. Book 5.1: 1-10, 22-25, 27-36, 40-46
5. Handout 5510
6. Book 5.3 page 260: 11-13, 37-39
7. Book 5.4: 1, 3, 4, 16, 18, 42, 43
8. Handout 5520
9. Book 5.5: 1, 4, 7, 10, 13, 16, 19, 21, 22, 24, 25, 34, 35
10. Book 5.R: 23, 24, 26, 27, 29, 30-42 page 284: 5, 6
A rational function, in lowest terms, will have vertical asymptotes at the real zeros of the denominator.
Examples
This above example has the zero 3 in the denominator. The vertical line x = 3 is a boundary for this rational function. Mentally, do you see the line x = 3. It is a boundary, It is an asymptote. This graph also has a horizontal boundary or asymptote. More about that in a few moments.
This above example has two zeros in the denominator: -3 and 4. The vertical lines x = -3 and x = 4 are vertical asymptotes. The curve dares not to touch these vertical asymptotes.
This above example has two zeros in the denominator: 0 and -2. You did see that 0 is a double root, Right? The vertical lines x = 0 and x = -2 are vertical asymptotes. Do you see how the curve rises on both sides of the vertical asymptote x = 0. Because it was a double root, this is a characteristic of that. The curve will either run up or down each side of an asymptote that was a double root or more exactly a root that occurred an even amount of times.
We find the vertical asymptotes by solving a rational function denominator for zero. These also correspond to the domain restrictions for the function.
Vertical asymptotes are sacred ground, horizontal asymptotes are just useful suggestions. You can never touch a vertical asymptote, you can (and often do) touch and even cross horizontal asymptotes.
1. If the degree of the numerator is less than the degree of the denominator, the rational function is a proper rational function and will have the horizontal asymptote y = 0.
Examples
In the above example, the degree of the numerator is less than the degree of the denominator. When this happens, the rational function is proper and the horizontal asymptote is the horizontal line y = 0. The horizontal line y = 0 is a horizontal boundary and in this example it is not very strict — the middle curve crosses the horizontal asymptotes. It will sometimes do this with horizontal asymptotes, never a vertical asymptote. We also have the vertical asymptotes x = 0 and x = -4.
In the above example, the degree of the numerator is also less than the degree of the denominator. Again when this happens, the rational function is proper and the horizontal asymptote is the horizontal line y = 0. The horizontal line y = 0 is a horizontal boundary. We also have the vertical asymptotes x = 1 and x = -2.
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Is this a proper rational function? Why? Yes, the degree of the numerator is less than the degree of the denominator.
The horizontal asymptote is y = 0.
The vertical asymptote is x = -2. And do you see that the curve is rising on both sides of this asymptote. This is indicative of the fact that -2 was a double root. When that happens the curve will either rise or fall on both sides of the vertical asymptote.
The degree of the numerators in these examples is less than the degree of the denominators. When this happens, the rational function is proper and the horizontal asymptote is the horizontal line y = 0.
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2. If the degree of the numerator is greater than or equal to the degree of the denominator, the rational function is improper.
a. If the degree of the numerator is equal to the degree of the denominator, the quotient of the lead coefficients of the numerator and denominator is a horizontal asymptote.
Examples
In the above example, the degree for the numerator and denominator are both first. The horizontal asymptote is equal to the quotient of the lead coefficients. y = 2/1 = 2.
In the above example, the degrees for the numerator and denominator are both second. The horizontal asymptote is equal to the quotient of the lead coefficients. y = 1/1 = 1.
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b. If the degree of the numerator is exactly one more than the degree of the denominator, the rational function has a slant asymptote that you will derive by division techniques.
Examples
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In these two examples, the degree of the numerator is exactly one more than the degree of the denominator. Because of this, use division techniques, either long or synthetic division, to find the slant asymptote. Disregard any remainders.
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c. If the degree of the numerator is greater than one more than the degree of the denominator, the rational function has neither a horizontal nor a slant asymptote.
Since the degree of the numerator is more than one more than the denominator, there is no horizontal asymptote.
