Targeted Review Solutions

Group the words below by measure of center and measure of spread.
mode, range, mean, standard deviation, median

measures of center: mean, median, mode
measures of spread: range, standard deviation

Calculate the measures of center for: $\{15, 20, 25, 25, 30, 35, 40, 50, 75, 90\}$
Identify the distribution.

$Mean = 40.5 Median = 32.5 Mode : 25$
Median < Mean
Right-skewed

Calculate the standard deviation. Show your work.
$\{2, 3, 4, 4, 5, 5, 6, 7\}$

$μ = \frac{2 + 3 + 4 + 4 + 5 + 5 + 6 + 7}{8} = 4.5 \begin{matrix} {(x - μ)}^{2} = \\ {(2 - 4.5)}^{2} = 6.25 & \sqrt{σ^{2}} = \sqrt{\frac{18}{8}} \\ {(3 - 4.5)}^{2} = 2.25 & \begin{matrix} σ = \sqrt{2.25} \\ σ = 1.5 \end{matrix} \\ {(4 - 4.5)}^{2} = 0.25 \\ {(4 - 4.5)}^{2} = 0.25 \\ {(5 - 4.5)}^{2} = 0.25 \\ {(5 - 4.5)}^{2} = 0.25 \\ {(6 - 4.5)}^{2} = 2.25 \\ {(7 - 4.5)}^{2} = 6.25 \\ \sum_{} {(x - μ)}^{2} = 18 \end{matrix}$

$σ = 1.5$

Determine the approximate z-value for 37% of a standard normal distribution.

$z = - 0.33$

For problems 5–6, use the graph.

Explain why the graph is a polynomial function. Then state the end behavior.

This graph is a polynomial because it is a smooth, continuous graph.
$a = postive, n = odd x \to - \infty, f (x) \to - \infty, and x \to + \infty, f (x) \to + \infty$

Given the graph, name the possible number of real roots.

C = single root, B = double root
$1 + 2 + 1 + 1 = 5$

There are 5 possible real roots for the polynomial.

Solve with common logs. Round to the ten-thousandth.
$12^{(3 x - 2)} = 4$

$\begin{array}{rcl} \log 12^{(3 x - 2)} & = & \log 4 \\ (3 x - 2) \log 12 & = & \log 4 \\ 3 x - 2 & = & \frac{\log 4}{\log 12} \\ 3 x & = & \frac{\log 4}{\log 12} + 2 \\ x & = & \frac{1}{3} (\frac{\log 4}{\log 12} + 2) = 0.85262 \end{array}$

$x = 0.8526$

Solve with natural logs.
$(e^{x} - 1) (e^{x} - 3) = 0$

$\begin{matrix} e^{x} = 1 & e^{x} = 3 \\ \ln e^{x} = \ln 1 & \ln e^{x} = \ln 3 \\ x = \ln 1 & x = \ln 3 \\ x = 0 & x = 1.0986 \end{matrix}$

$x = 0, 1.0986$

Multiple Choice

Identify the type of distribution shown in the sketch.

Left-skewed
Right-skewed
Symmetric
Cannot be determined

$median < mean Right - skewed$

Note

A, C) These options do not represent the sketch shown.

Which equation is equivalent to $\ln 9 + 5 \ln x = 3 \ln 7$ ?

$\begin{array}{rcl} \ln 9 x^{5} & = & \ln 21 \end{array}$
$\begin{array}{rcl} \ln 9 x^{5} & = & \ln 343 \end{array}$
$\begin{array}{rcl} \ln 45 x & = & \ln 343 \end{array}$
$\ln 14 x = \ln 10$

Note

On the right side of the equation, 3 was not used as an exponent.

On the left side of the equation, 9 and 5 were multiplied together.
The numbers on each side were added together, ignoring the properties of logs.

$\begin{array}{rcl} \ln 9 + \ln x^{5} & = & \ln 7^{3} \\ \ln 9 x^{5} & = & \ln 343 \end{array}$

If $y = 15 {(0.5)}^{x},$ what is the approximate value of x when $y = 3$ ?

–2.32
0.55
0.43
2.32

$\begin{array}{rcl} \frac{3}{15} & = & \frac{15 {(0.5)}^{x}}{15} \\ 0.2 & = & 0 . 5^{x} \\ \log 0.2 & = & \log 0 . 5^{x} \\ \log 0.2 & = & x \log 0.5 \\ x & = & \frac{\log 0.2}{\log 0.5} \end{array}$

Note

This option occurs when 15 is divided by 3 in the first step.
This option occurs when 15 and 0.5 are multiplied, ignoring the properties of exponents.
This option occurs when the reciprocal of the last step is taken.

Determine the raw data value when the proportion of area for a standard normal distribution is 97% with a mean of 5 and a standard deviation of 2.1.

6.95
8.95
7.04
11.50

$\begin{matrix} 0.97 \approx 0.9699 z = 1.88 \end{matrix} 1.88 = \frac{X - 5}{2.1} 3.948 = X - 5 X = 8.948$

Note

The value 0.97 was used instead of the z-score, and the mean and standard deviation were switched in the formula.

The value 0.97 was used instead of the z-score.
The mean and standard deviation were switched in the formula.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L45	L45	L45	L46	L33	L35	L41	L42	L45	L41	L38	L46

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics, FD = Foundational Knowledge