Step-by-step explanation:
x(2)y(4)z(5)xy(2)

Answer:
The given statement describe inferential statistics.
Step-by-step explanation:
Descriptive Statistic:
- It helps us to summarize a given data set.
- It could describe the entire population or a sample from the population.
- There are two types of descriptive measures: measures of central tendency and measures of variability
- Central Tendency: Mean, mode, Median
- Measure of Viability: Standard Deviation, Variance, Range, Interquartile range
Inferential Statistic:
- It s the process of estimating population parameter with the help of a sample from the population.
- A random sample from the population is used to describe the population with the help of sample statistic.
Given Scenario:
"Based on a sample of 170 truck drivers, there is evidence to indicate that, on average, independent truck drivers earn more than company-hired truck drivers."
Thus, this is an example of a inferential statistics as a sample was used to estimate the population.
Here,
Sample:
Sample of 170 truck drivers
Population:
All truck drivers.
With the help of a sample, we approximated the population, thus, this statement describe inferential statistics.
QR is congruent to PS
So if QR is 17.7 PS is 17.7
Answer 3.) PS equals 17.7
Answer:
<h2>
£1,330.46</h2>
Step-by-step explanation:
Using the compound interest formula 
A = amount compounded after n years
P = principal (amount invested)
r = rate (in %)
t = time (in years)
n = time used to compound the money
Given P = £1200., r = 3.5%, t = 3years, n = 1 year(compounded annually)

Value of Charlie's investment after 3 years is £1,330.46