a.
Tocalculate:The mean, median and mode of the prices if the ad says “Our flat-screen TVs average $695. The prices of the flat screen TVs are $1200, $999, $1499, $895, $695, $1100, $1300 and $695.
The mean price is $1047.875. The median price is $1049.5. the mode of the data set is $695.
Given information:
The ad says “Our flat-screen TVs average $695. The prices of the flat screen TVs are $1200, $999, $1499, $895, $695, $1100, $1300 and $695.
Formula used:
Calculation:
Consider the given information.
Mean:
The mean price is $1047.875
Median:
Median is the middle value of the ordered data set.
There are 8 values in the data set so median will be the average of 4th and 5th data value.
The median price is $1049.5
Mode is the highest frequency value of the given data set. It can be observed that the frequency of 695 is 2, which is the highest among all.
So, the mode of the data set is $695.
b.
To identify:The measure which used by the store in its ad? Why did they choose it?
The store used mode in the ad. They choose mode because it is the lowest among mean, mode and median.
Given information:
The mean price is $1047.875. The median price is $1049.5. the mode of the data set is $695.
Explanation:
Consider the given information.
The mean price is $1047.875. The median price is $1049.5. the mode of the data set is $695.
The ad says “Our flat-screen TVs average $695.
Thus, store used mode in the ad. They choose mode because it is the lowest among mean, mode and median.
c.
To identify:Which measure should be used by the consumer.
Median should be used by the consumer.
Given information:
The mean price is $1047.875. The median price is $1049.5. the mode of the data set is $695.
The ad says “Our flat-screen TVs average $695.
Explanation:
Consider the given information.
As a consumer median is the measure of the average which is least affected by the outliers and skewness of the distribution.
So, choose median as the average.
Chapter 11 Solutions
High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11
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- For the following exercise, find the domain and range of the function below using interval notation. 10+ 9 8 7 6 5 4 3 2 1 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 2 34 5 6 7 8 9 10 -1 -2 Domain: Range: -4 -5 -6 -7- 67% 9 -8 -9 -10-arrow_forward1. Given that h(t) = -5t + 3 t². A tangent line H to the function h(t) passes through the point (-7, B). a. Determine the value of ẞ. b. Derive an expression to represent the gradient of the tangent line H that is passing through the point (-7. B). c. Hence, derive the straight-line equation of the tangent line H 2. The function p(q) has factors of (q − 3) (2q + 5) (q) for the interval -3≤ q≤ 4. a. Derive an expression for the function p(q). b. Determine the stationary point(s) of the function p(q) c. Classify the stationary point(s) from part b. above. d. Identify the local maximum of the function p(q). e. Identify the global minimum for the function p(q). 3. Given that m(q) = -3e-24-169 +9 (-39-7)(-In (30-755 a. State all the possible rules that should be used to differentiate the function m(q). Next to the rule that has been stated, write the expression(s) of the function m(q) for which that rule will be applied. b. Determine the derivative of m(q)arrow_forwardSafari File Edit View History Bookmarks Window Help Ο Ω OV O mA 0 mW ర Fri Apr 4 1 222 tv A F9 F10 DII 4 F6 F7 F8 7 29 8 00 W E R T Y U S D பட 9 O G H J K E F11 + 11 F12 O P } [arrow_forward
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