Question

1. select all the equations that have the same solution as 2x - 5 = 15. a. 2x = 10 b. 2x = 20 c. 2(x - 5) = 15 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 5.82 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 set: a. what is the mean? b. what is the standard deviation? c. what is the median? d. what is the iqr? (from unit 1, lesson 14.)

Explanation:

Step1: Solve the equation $2x - 5=15$

Add 5 to both sides: $2x-5 + 5=15 + 5$, so $2x=20$. Then divide both sides by 2, $x = 10$.

Step2: Check option A

For $2x = 10$, divide both sides by 2, $x = 5
eq10$.

Step3: Check option B

For $2x = 20$, divide both sides by 2, $x = 10$.

Step4: Check option C

For $2(x - 5)=15$, distribute the 2: $2x-10 = 15$. Add 10 to both sides: $2x=25$, then $x=\frac{25}{2}
eq10$.

Step5: Check option D

For $2x-20 = 0$, add 20 to both sides: $2x=20$, then $x = 10$.

Step6: Check option E

For $4x-10 = 30$, add 10 to both sides: $4x=40$. Divide both sides by 4, $x = 10$.

Step7: Check option F

For $15=5 - 2x$, subtract 5 from both sides: $10=-2x$. Divide both sides by - 2, $x=-5
eq10$.

for question 5:

Step1: Recall the effect of an out - lier on mean

The mean is affected by outliers. Let the sum of all data points be $S$ and the number of data points be $n$. The mean $\bar{x}=\frac{S}{n}=5$. Let the outlier value be $x_{o}=26$. After removing the outlier, the new sum is $S - 26$ and the new number of data points is $n - 1$. Since the outlier is a large value, the new mean will be less than 5. But without knowing the exact number of data - points, we assume the sum of the non - outlier data points is $S-26$. If we assume there were $n$ data points originally and the mean $\bar{x}=\frac{S}{n}=5$, so $S = 5n$. The new mean $\bar{x}_{new}=\frac{5n - 26}{n - 1}$. Since we can't count the dots precisely from the image, we note that the non - outlier data is more concentrated around lower values. The new mean will be closer to the median. Since the median of non - outlier data is 4, the new mean will be less than 5 and close to 4. Let's assume there are 20 data points originally (by counting the dots approximately). $S=5\times20 = 100$, after removing 26, the new sum is $100 - 26=74$ and new number of data points is 19. The new mean is $\frac{74}{19}\approx3.89$. But a more intuitive way is to note that the non - outlier data is clustered around 4, so the new mean is approximately 4.

Step2: Recall the effect of an out - lier on standard deviation

The standard deviation measures the spread of the data. The outlier 26 increases the spread of the data. Removing it will decrease the standard deviation. The original standard deviation is 5.82. After removing the outlier, the data is more clustered together, so the new standard deviation will be less than 5.82. A rough estimate by looking at the non - outlier data's spread is around 2 (by observing the distance of data points from the median).

Step3: Recall the effect of an out - lier on median

The median is the middle value of the data set. Since the outlier is an extreme value (26) and we assume the number of non - outlier data points is even or odd, the position of the middle value among non - outlier data is not affected much by the removal of the outlier. The original median of non - outlier data is 4, so the new median is 4.

Step4: Recall the effect of an out - lier on IQR

The inter - quartile range (IQR) is the difference between the third quartile and the first quartile. The outlier 26 is an extreme value and is not part of the middle 50% of the data. Removing it will not change the IQR. The original IQR is 3, so the new IQR is 3.

Answer:

B. $2x = 20$
D. $2x - 20 = 0$
E. $4x - 10 = 30$