Question

1. select all the equations that have the same solution as 2x - 3 = 13. a. 2x = 16 b. 2x = 20 c. 2(x - 3)=13 d. 2x - 20 = 0 e. 4x - 10 = 30 f. 15 = 5 - 2x 5. the number of hours spent in an airplane on a single flight is recorded on a dot - plot. the mean is 5 hours and the standard deviation is approximately 3.82 hours. the median is 4 hours and the iqr is 3 hours. the value 26 hours is an outlier that should not have been included in the data. when the outlier is removed from the data: a. what is the mean? b. what is the standard deviation? c. what is the median? d. what is the iqr?

Explanation:

Step1: Solve the equation \(2x - 5=15\)

Add 5 to both sides: \(2x=15 + 5=20\), then \(x = 10\).

Step2: Check option A \(2x=10\)

Solve for \(x\): \(x = 5
eq10\).

Step3: Check option B \(2x = 20\)

Solve for \(x\): \(x = 10\).

Step4: Check option C \(2(x - 5)=15\)

Expand: \(2x-10 = 15\), then \(2x=25\), \(x=\frac{25}{2}
eq10\).

Step5: Check option D \(2x-20 = 0\)

Add 20 to both sides: \(2x=20\), \(x = 10\).

Step6: Check option E \(4x-10 = 30\)

Add 10 to both sides: \(4x=40\), \(x = 10\).

Step7: Check option F \(15=5 - 2x\)

Add \(2x\) to both sides: \(2x + 15=5\), then \(2x=- 10\), \(x=-5
eq10\).

For question 5:
Let the sum of all data - points be \(S\) and the number of data - points be \(n\). We know that \(\bar{x}=\frac{S}{n}=5\). Let's assume the outlier value is \(x_{o}=26\).

Step1: Calculate the new mean

If the mean \(\bar{x}=\frac{S}{n}=5\), then \(S = 5n\). After removing the outlier, the new sum is \(S'=S - 26\) and the new number of data - points is \(n'=n - 1\). We don't know \(n\), but we can use the concept in a general way. Since we don't have the exact number of data - points, assume there are \(k\) non - outlier data points. Let the sum of non - outlier data points be \(S_{1}\). We know that \(S=S_{1}+26\) and \(\frac{S}{n}=5\). If we assume the original number of data points is \(n\), then \(S = 5n\). After removing the outlier, the new mean \(\bar{x}'=\frac{5n - 26}{n - 1}\). Without knowing \(n\), if we assume we have enough data points, we can also note that the outlier has a large impact. But if we assume for simplicity that we can work with the fact that the sum of all points except the outlier is \(S_{1}\) and we know that the mean of all points is 5. Let's assume there are \(n\) points originally. So \(S = 5n\). After removing the outlier, the new sum \(S'=5n-26\) and new number of points \(n'=n - 1\). Since the outlier is large, the new mean will be less than 5. But if we assume we have 10 data points originally (\(n = 10\), \(S=50\)), after removing the outlier \(S'=50 - 26=24\) and \(n'=9\), the new mean \(\bar{x}'=\frac{24}{9}\approx2.67\). In a more general sense, we know that the mean is affected by outliers. The new mean will be closer to the central tendency of the non - outlier data. Since the outlier is large, the new mean will be less than 5. Let's assume the data set has \(n\) values with sum \(S\) such that \(\frac{S}{n}=5\) or \(S = 5n\). After removing the outlier \(x = 26\), the new sum is \(S'=S - 26=5n-26\) and the new number of data points is \(n'=n - 1\). The new mean \(\bar{x}=\frac{5n-26}{n - 1}\).
The standard deviation measures the spread of the data. Removing a large outlier will decrease the standard deviation. The original standard deviation is approximately \(5.82\). After removing the outlier, the data is more clustered around the central values, so the new standard deviation will be less than \(5.82\).
The median is 4. Removing a single outlier (especially when the number of data points is relatively large) may not change the median if the outlier is at one of the extremes. Since the median is 4 and the outlier is 26 (a large value), the median will remain 4.
The inter - quartile range (IQR) is 3. Removing a single outlier at the upper end of the data will not change the IQR if the outlier is above the third quartile. So the IQR will remain 3.

Answer:

1. B. \(2x = 20\), D. \(2x-20 = 0\), E. \(4x-10 = 30\)

5.
a. Less than 5 (exact value depends on the number of data - points, but conceptually it will be pulled down by removing the large outlier)
b. Less than \(5.82\)
c. 4
d. 3