18. attention span\n\nscatter - plot showing ...
18. attention span\n\nscatter - plot showing age (years) on x - axis and time (minutes) on y - axis\n\na. no correlation\nb. positive correlation\nc. negative correlation\nd. weak correlation\n\n19. find the indicated function value: (f(5)) for (f(x)=x^{2}-2x + 1).\na. 16\nb. 1\nc. 5\nd. does not exist\n\n20. the mass of the moon is about (7.348\times10^{22}) kilograms and the mass of earth is (5.972\times10^{24}) kilograms. how many times greater is earths mass than the moons mass?\na. (8.127\times10^{1})\nb. 8.127\nc. 812.7\nd. (8.127\times10^{-1})
Answer
### 18.
# Explanation:
## Step1: Observe the scatter - plot
As age increases, the attention - span (time in minutes) generally increases.
## Step2: Recall correlation types
Positive correlation means as one variable increases, the other variable also increases.
# Answer:
b. Positive correlation
### 19.
# Explanation:
## Step1: Substitute \(x = 5\) into the function
Given \(f(x)=x^{2}-2x + 1\), substitute \(x = 5\): \(f(5)=5^{2}-2\times5 + 1\).
## Step2: Calculate the result
\[
\begin{align*}
f(5)&=25-10 + 1\\
&=16
\end{align*}
\]
# Answer:
a. 16
### 20.
# Explanation:
## Step1: Set up the ratio
To find how many times Earth's mass is greater than the Moon's mass, divide Earth's mass by the Moon's mass. Let \(M_E = 5.972\times10^{24}\) kg and \(M_M=7.348\times10^{22}\) kg. The ratio is \(\frac{M_E}{M_M}=\frac{5.972\times10^{24}}{7.348\times10^{22}}\).
## Step2: Use exponent rules
\(\frac{5.972\times10^{24}}{7.348\times10^{22}}=\frac{5.972}{7.348}\times10^{24 - 22}\).
\(\frac{5.972}{7.348}\approx0.8127\), and \(10^{24 - 22}=10^{2}\). So \(0.8127\times10^{2}=81.27 = 8.127\times10^{1}\).
# Answer:
a. \(8.127\times10^{1}\)