Task 1: Normal Distribution Scores On A Statewide Standardiz

Task 1normal Distributionscores On A Statewide Standardized Test Are

On a statewide standardized test, scores are normally distributed with a mean of 12.89 and a standard deviation of 1.95. Certificates are awarded to students whose scores are in the top 2% of test takers, meaning they scored better than 98% of others. Marcus scored 13.7 and is wondering if he qualifies for a certificate. To determine this, we need to find the score corresponding to the 98th percentile of the distribution and compare it to Marcus's score.

To begin, we calculate the z-score that marks the top 2% of a normal distribution. Using standard normal distribution tables or a calculator, the z-score corresponding to the 98th percentile is approximately 2.05. This value indicates that 98% of scores are below a z-score of 2.05.

Next, we convert this z-score back to the original score scale using the formula:

X = μ + zσ

where μ is the mean (12.89), σ is the standard deviation (1.95), and z is 2.05. Calculating, we find:

X = 12.89 + 2.05 * 1.95 ≈ 12.89 + 4.00 = 16.89

This means that the score needed to be in the top 2% is approximately 16.89. Since Marcus scored 13.7, which is significantly below 16.89, he did not meet the threshold for receiving a certificate.

Paper For Above instruction

Dear Marcus,

I am writing to inform you about whether your test score qualified you for a certificate based on the top 2% criterion. The analysis involved understanding the distribution of scores on the standardized test, which follows a normal distribution with a mean score of 12.89 and a standard deviation of 1.95.

To determine the cutoff score for the top 2%, I calculated the z-score corresponding to the 98th percentile of the standard normal distribution, which is approximately 2.05. Using this z-score, I converted back to the original score scale using the formula X = μ + zσ. This resulted in a cutoff score of about 16.89. Consequently, any student scoring above this value would be in the top 2% and eligible for a certificate.

Since your score was 13.7, which is below the cutoff of approximately 16.89, you did not qualify for the certificate. This analysis shows that while your score is above the average, it did not reach the performance level required for the top 2% distinction. Keep studying and striving, and you may be able to achieve this in future assessments.

Sincerely,

[Your Name]

Task 2—Non-Linear Systems of Equations

Let’s define the system of equations. For the linear equation, we choose:

y = 2x + 3

For the quadratic equation, we select:

y = x² - 1

Part 1: Algebraic Solution

To solve the system, set the two equations equal to each other:

2x + 3 = x² - 1

Bring all terms to one side:

x² - 2x - 4 = 0

This quadratic equation can be solved using the quadratic formula:

x = [2 ± √(4 - 4 1 (-4))]/2

= [2 ± √(4 + 16)]/2

= [2 ± √20]/2

= [2 ± 2√5]/2

= 1 ± √5

Thus, the solutions for x are:

x = 1 + √5 ≈ 1 + 2.236 = 3.236

x = 1 - √5 ≈ 1 - 2.236 = -1.236

Substitute these x-values back into y = 2x + 3 to find y:

When x ≈ 3.236, y ≈ 2(3.236) + 3 ≈ 6.472 + 3 = 9.472

When x ≈ -1.236, y ≈ 2(-1.236) + 3 ≈ -2.472 + 3 = 0.528

Part 2: Graphical Solution

Plotting both functions:

• y = 2x + 3 (a line with slope 2 and y-intercept 3)
• y = x² - 1 (a parabola opening upwards with vertex at (0, -1))

The intersection points correspond to the solutions found above, approximately at (3.236, 9.472) and (-1.236, 0.528). The graph visually confirms these solutions, illustrating where the line and parabola intersect.

Task 3—Graphing Rational Functions

Let’s define the rational function:

f(x) = (2x + 1)/(x - 3)

Part 1: Graphing the Function

Using graphing technology such as Desmos or GeoGebra, the visual graph reveals the vertical asymptote at x = 3 (where the denominator zeroes out), and the horizontal asymptote at y = 2 (since as x approaches infinity, f(x) approaches the ratio of leading coefficients).

The x-intercept occurs where the numerator equals zero: 2x + 1 = 0 => x = -0.5, y = 0.

The y-intercept is at x = 0: f(0) = (2*0 + 1)/(0 - 3) = 1/(-3) = -1/3.

Part 2: Identifying Asymptotes and Intercepts

• Vertical asymptote: x = 3, where the denominator equals zero.
• X-intercept: x = -0.5, f(x)=0.
• Y-intercept: at (0, -1/3).

Task 4—Solving Rational Equations with Extraneous Solutions

Given the form: (x + a)/(a x) = b/x, we select specific values for a and b that lead to extraneous solutions. Choose a = 2, b = 4.

The equation becomes:

(x + 2)/(2x) = 4/x

Part 1: Solve the Equation

Cross-multiplied:

(x + 2) x = 4 2x

x(x + 2) = 8x

x² + 2x = 8x

x² + 2x - 8x = 0

x² - 6x = 0

x(x - 6) = 0

Solutions: x = 0 or x = 6

However, check for extraneous solutions by substituting back into the original equation. At x=0, the left side is undefined because of division by zero, so discard x=0. The valid solution is x=6.

Part 2: Explanation of Extraneous Solutions

The extraneous solution arises from algebraic manipulations that introduce invalid solutions—such as values that make denominators zero—so verification against the original equation is essential. Here, x=0 is extraneous because it makes the denominator zero, and solutions must be checked to confirm their validity.

Task 5—Polynomial Division and the Remainder Theorem

Create a quadratic polynomial: f(x) = x² + 3x + 2

Choose a linear binomial: (x - 1)

Part 1: Polynomial Division

Divide f(x) = x² + 3x + 2 by (x - 1):

Using long division:

• Divide x² by x: x
• Multiply (x - 1) by x: x² - x
• Subtract: (x² + 3x + 2) - (x² - x) = 4x + 2
• Divide 4x by x: 4
• Multiply (x - 1) by 4: 4x - 4
• Subtract: (4x + 2) - (4x - 4) = 6

So, the quotient is x + 4, and the remainder is 6.

Part 2: Evaluate f(1):

f(1) = 1² + 3(1) + 2 = 1 + 3 + 2 = 6

This matches the remainder, illustrating the Remainder Theorem, which states that the remainder when dividing f(x) by (x - a) equals f(a). Since the remainder equals f(1), (x - 1) is not a factor of f(x) because the remainder is not zero.

Task 6—Polynomial Identities

Choose a two-digit number greater than 25, for example, 36.

Rewrite 36 as a difference of two numbers:

36 = 25 + 11

Express 36 as a difference: 36 = 50 - 14

Use the identity: (x - y)^2 = x^2 - 2xy + y^2

Let’s take x = 36. Rewrite as:

36 = (25 + 11)

Square 36 using the identity:

(x)^2 = (25 + 11)^2 = 25^2 + 2 25 11 + 11^2 = 625 + 550 + 121 = 1296

Alternatively, expressing the number as 50 - 14:

(50 - 14)^2 = 50^2 - 2 50 14 + 14^2 = 2500 - 1400 + 196 = 1296

This approach demonstrates how to square a specific two-digit number without a calculator by deconstructing it into simpler components and applying the difference of squares and the identity above.

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