SAT Randomized Questions - 1 Full Math Practice Test - Answers and Detailed Explanations at the END

Data set A is skewed toward lower values, while data set B is skewed toward higher values. Therefore, the mean of Data set A is less than the mean of Data set B.

Since the function decreases by 30%, it retains 70% of its value at each step. Thus, the multiplier for the decay is 0.70. Therefore, the correct equation is \( f(x) = 100(0.70)^x \).

Since \(g(x) = f(x+2) = 8(5)^{x+2} = 8 \cdot 25 \cdot (5)^x = 200(5)^x\), the correct answer is g(x) = 200(5)^x.

Start with the slope \( -\frac{5}{6} \) and point \( (-6, 4) \). Using the point-slope formula, the equation is \( y = -\frac{5}{6}x + 4 \).

For a quadratic in the form \( y = ax^2 + bx + c \), the minimum or maximum value of \( y \) occurs at \( x = \frac{-b}{2a} \). Here, \( a = 1 \) and \( b = -10 \), so \( x = \frac{10}{2} = 5 \).

First, find the measure of angle T in radians: \( \frac{2\pi}{3} + \frac{\pi}{4} = \frac{11\pi}{12} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{11\pi}{12} \cdot \frac{180}{\pi} = 165 \) degrees.

The minimum value of \( y \) occurs at \( x = \frac{-b}{2a} \). Here, \( a = 1 \) and \( b = -8 \), so \( x = \frac{8}{2} = 4 \).

Since \( m(3) = 4096 \), substitute into \( m(x) = 16 \cdot q^x \): \( 16 \cdot q^3 = 4096 \Rightarrow q^3 = 256 \Rightarrow q = 4 \). Now find \( m(\frac{1}{4}) \): \( m(\frac{1}{4}) = 16 \cdot 4^{\frac{1}{4}} = 16 \cdot \sqrt[4]{4} = 16 \cdot \sqrt{2} = 16 \sqrt{2} \).

Substitute values from each answer choice into both equations. Only \( (5, 10) \) satisfies both equations.

Let the number of small candles be represented by \( x \), and the number of large candles by \( y \). The total number of candles is at least 200, so we have the inequality:

\( x + y geq 200 \).

The budget constraint is that the total cost must not exceed $2,200. Therefore, we have the equation:

\( 4.90x + 11.60y leq 2,200 \).

We want to maximize the number of large candles, \( y \). First, solve for \( x \) in terms of \( y \) using the inequality:

\( x = 200 - y \).

Substitute this into the budget constraint:

\( 4.90(200 - y) + 11.60y leq 2,200 \)

Simplify:

\( 980 - 4.90y + 11.60y leq 2,200 \)

Combine like terms:

\( 980 + 6.70y leq 2,200 \)

Subtract 980 from both sides:

\( 6.70y leq 1,220 \)

Solve for \( y \):

\( y leq rac{1,220}{6.70} approx 182.09 \)

Since the number of candles must be an integer, the maximum number of large candles is 182.

Solving for \( y \) using substitution or elimination, we find that \( y = 30 \).

In the function g(x) = -3x + 50, the y-intercept (where x = 0) represents the initial amount of dough the baker had before making any loaves. When x = 0, g(0) = 50, indicating that the baker started with 50 pounds of dough.

Substitute \( y = 24 \) into the equation: \( 3.5x + 7(24) = 140 \Rightarrow 3.5x + 168 = 140 \Rightarrow 3.5x = -28 \). Since \( x \) cannot be negative, the shelter cannot care for any cats, making this question tricky.

Testing each pair shows that only \( (4, 5) \) satisfies both equations.

We use the Pythagorean theorem: \( c^2 = 9^2 + 12^2 = 81 + 144 = 225 \). Therefore, \( c = \sqrt{225} = 15 \). The hypotenuse can be written as \( 3\sqrt{25} \), so \( d = 25 \).

The decay factor is 0.90 per three months, equating to a 10% drop per quarter. After one year, this approximates to a 34% annual decay rate, so \(m = 34\).

Convert the area to square miles by dividing by \(1760^2\): \(10,240,000 / 3,097,600 = 3.31\) square miles.

In the function f(x) = -5x + 30, the y-intercept (where x = 0) represents the initial amount of juice Caleb had before he made any popsicles. When x = 0, f(0) = 30, indicating that Caleb started with 30 fluid ounces of juice.

The decay factor is 0.75 per half-year, which means it decays by 25% every six months. Over a year, this represents a 25% drop, so \(q = 25\).

Calculate the slope with the given points and determine the equation of the line. Substitute \(x = p - 6\) to find \(b = -41\).

Since \(g(x) = f(x+1) = 6(10)^{x+1} = 6 \cdot 10 \cdot (10)^x = 60(10)^x\), the correct answer is g(x) = 60(10)^x.

Only \( (2, 6) \) satisfies both equations.

The population grows by 2% annually, so the growth factor is 1.02. Since \( x \) is in seconds, we convert it to years by dividing by \( 365 \cdot 24 \cdot 60 \cdot 60 \). The equation is \( p = 50,000 \cdot 1.02^{\frac{x}{365 \cdot 24 \cdot 60 \cdot 60}} \).

The percentage can be calculated using the formula: \( \frac{\text{Part}}{\text{Whole}} \times 100 \). Here, \( \frac{50}{200} \times 100 = 25\% \).

Both equations are identical, which means the lines coincide perfectly and intersect at infinitely many points.

Using \( x = \frac{-b}{2a} \), we find the minimum value occurs at \( x = \frac{12}{2} = 6 \).

First, find the measure of angle N in radians: \( \frac{3\pi}{4} + \frac{\pi}{6} = \frac{11\pi}{12} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{11\pi}{12} \cdot \frac{180}{\pi} = 165 \) degrees.

The decay factor is 0.68 every half-year, which means it decays by 32% each six months. This translates to approximately a 55% annual decay rate, so \(p = 55\).

To find \( a \), set \( p(x) = 0 \): \( 3x + 9 = 0 Rightarrow x = -3 \). Thus, \( a = -3 \). For \( b \), evaluate \( p(0) = 9 \). Therefore, \( a + b = -3 + 9 = 6 \).

Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 28}}{2} = 4 \pm \sqrt{9}\). So, the solution can be written as \(4 + \sqrt{9}\) and the correct value of \(k\) is 9.

To find \( a \), set \( f(x) = 0 \) and solve for \( x \): \( 5x - 15 = 0 Rightarrow x = 3 \). The x-intercept is \( (3, 0) \), so \( a = 3 \). The y-intercept \( b \) is found by setting \( x = 0 \): \( f(0) = -15 \), so \( b = -15 \). Then \( a + b = 3 - 15 = -12 \).

With \( h(4) = 810 \), substitute into \( h(x) = 10 \cdot c^x \): \( 10 \cdot c^4 = 810 \Rightarrow c^4 = 81 \Rightarrow c = 3 \). Now find \( h(-1) \): \( h(-1) = 10 \cdot 3^{-1} = 10 \cdot \frac{1}{3} = \frac{10}{3} \).

Congruent triangles have corresponding angles that are equal. Since \( L \) and \( O \) correspond, angle \( O \) must be 40°.

The two equations are actually multiples of each other. Dividing the second equation by 2 yields \( 3x + 5y = 25 \). Thus, both lines coincide, which means they intersect at infinitely many points.

After simplifying and solving, we find \( y = 27 \).

Calculate percentages: Candidate L received \( \frac{600}{1000} = 0.6 \), or 60%. Candidate M received \( \frac{400}{1000} = 0.4 \), or 40%. If 7,500 people vote, Candidate L would receive \( 0.6 \times 7500 = 4500 \) votes, and Candidate M \( 0.4 \times 7500 = 3000 \) votes. The difference is \( 4500 - 3000 = 1500 \) votes.

The percentage can be calculated using the formula: \( \frac{350}{700} \times 100 = 50\% \).

Since the side lengths of square M are 15 times those of square N, the area of square M is \( 15^2 = 225 \) times the area of square N. Therefore, \( k = 225 \).

Rewrite \(f(x)\) in vertex form as \(5(x - 2)^2 + 75\). The minimum of \(f(x)\) is at \(x = 2\), so \(g(x) = f(x + 3)\) shifts this to \(x = -1\).

Since \(g(x) = f(x+3) = 3(6)^{x+3} = 3 \cdot 216 \cdot (6)^x = 648(6)^x\), the correct answer is g(x) = 648(6)^x.

From \(\frac{x}{y} = 8\), we know \(x = 8y\). Substitute into \(\frac{56x}{my} = 8\) to get \(\frac{56(8y)}{my} = 8\). Simplify to \(\frac{448y}{my} = 8\), cancel \(y\) and solve to find \(m = 56\).

First, find the measure of angle Y in radians: \( \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{\pi}{2} \cdot \frac{180}{\pi} = 90 \) degrees.

With the side lengths of square R being 12 times those of square S, the area of square R is \( 12^2 = 144 \) times the area of square S. Thus, \( k = 144 \).

Data set B has higher frequencies around the center value (20) compared to Data set A, resulting in equal means for both data sets.