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John is trying to find the relationship between a student's age and the number of pets that he or she owns. He collects data from a random sample of students at his high school. He plots the data on a scatter plot.



What relationship between age and number of pets is John most likely to see?

Answer options with 4 options
A.
constant

B.
negative

C.
positive

D.
no correlation

3 answers

The relationship between a student's age and the number of pets they own can vary based on many factors, but generally, we might expect that as students age, they may have more responsibilities and possibly live in situations where they can have pets.

Given this context, it is reasonable to infer that:

  • Younger students (like those in middle school) might have fewer pets due to restrictions from parents or living arrangements.
  • Older students (like those in high school) might have more opportunities to own pets.

Based on these typical trends, the most likely relationship John is to see between a student's age and the number of pets they own is:

C. positive

Therefore, the answer is C. positive.

The area of a rectangle is 24 square centimeters. The length (x), in centimeters, of the rectangle is given by the equation x squared plus 5 x equals 24.



What is the length of the rectangle?

Answer options with 4 options
A.
3 centimeters

B.
5 centimeters​

C.
8 centimeters​

D.
12 centimeters

To find the length of the rectangle, we start with the given equation:

\[ x^2 + 5x = 24. \]

First, we rearrange this equation to standard quadratic form:

\[ x^2 + 5x - 24 = 0. \]

Now, we will factor this quadratic equation. We need two numbers that multiply to \(-24\) and add to \(5\). These numbers are \(8\) and \(-3\).

So, we can factor the equation as:

\[ (x + 8)(x - 3) = 0. \]

Setting each factor equal to zero gives us:

  1. \( x + 8 = 0 \) → \( x = -8 \) (not a valid length)
  2. \( x - 3 = 0 \) → \( x = 3 \)

Since length cannot be negative, we discard \( x = -8\).

Thus, the length of the rectangle is:

A. 3 centimeters.