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Answer :
Sure, let's solve both questions step-by-step.
Question 22:
Find the mean, median, and mode of the data: [tex]\(141, 144, 133, 128, 141, 136,146, 159, 143, 141\)[/tex].
1. Mean:
The mean is the average of all the numbers. To find the mean, add up all the numbers and then divide by the number of values.
Sum of the numbers:
[tex]\(141 + 144 + 133 + 128 + 141 + 136 + 146 + 159 + 143 + 141 = 1412\)[/tex]
Number of values:
[tex]\(10\)[/tex]
Mean:
[tex]\[
\text{Mean} = \frac{1412}{10} = 141.2
\][/tex]
2. Median:
The median is the middle number when the numbers are arranged in order.
First, arrange the numbers in ascending order:
[tex]\(128, 133, 136, 141, 141, 141, 143, 144, 146, 159\)[/tex]
Since there are 10 numbers (an even number), the median will be the average of the 5th and 6th numbers in this ordered list:
[tex]\[
\text{Median} = \frac{141 + 141}{2} = 141.0
\][/tex]
3. Mode:
The mode is the number that appears most frequently in the list.
From the list [tex]\(141, 144, 133, 128, 141, 136, 146, 159, 143, 141\)[/tex], the number [tex]\(141\)[/tex] appears most frequently (3 times).
Mode:
[tex]\[
\text{Mode} = 141
\][/tex]
Summary of Question 22:
- Mean: [tex]\(141.2\)[/tex]
- Median: [tex]\(141.0\)[/tex]
- Mode: [tex]\(141\)[/tex]
Question 23:
Evaluate if [tex]\(x = 7\)[/tex], [tex]\(y = 4\)[/tex], and [tex]\(z = 4\)[/tex]:
[tex]\[
\left(y^2 - x^2\right) \div (9z - 2)
\][/tex]
Substitute the given values into the expression:
[tex]\[
x = 7, \quad y = 4, \quad z = 4
\][/tex]
First, calculate [tex]\(y^2 - x^2\)[/tex]:
[tex]\[
y^2 = 4^2 = 16
\][/tex]
[tex]\[
x^2 = 7^2 = 49
\][/tex]
[tex]\[
y^2 - x^2 = 16 - 49 = -33
\][/tex]
Next, calculate [tex]\(9z - 2\)[/tex]:
[tex]\[
9z = 9 \cdot 4 = 36
\][/tex]
[tex]\[
9z - 2 = 36 - 2 = 34
\][/tex]
Finally, divide [tex]\((y^2 - x^2)\)[/tex] by [tex]\((9z - 2)\)[/tex]:
[tex]\[
\frac{y^2 - x^2}{9z - 2} = \frac{-33}{34} \approx -0.9705882352941176
\][/tex]
Summary of Question 23:
The value of the expression is approximately [tex]\(-0.9705882352941176\)[/tex].
I hope this helps! If you have any other questions, feel free to ask.
Question 22:
Find the mean, median, and mode of the data: [tex]\(141, 144, 133, 128, 141, 136,146, 159, 143, 141\)[/tex].
1. Mean:
The mean is the average of all the numbers. To find the mean, add up all the numbers and then divide by the number of values.
Sum of the numbers:
[tex]\(141 + 144 + 133 + 128 + 141 + 136 + 146 + 159 + 143 + 141 = 1412\)[/tex]
Number of values:
[tex]\(10\)[/tex]
Mean:
[tex]\[
\text{Mean} = \frac{1412}{10} = 141.2
\][/tex]
2. Median:
The median is the middle number when the numbers are arranged in order.
First, arrange the numbers in ascending order:
[tex]\(128, 133, 136, 141, 141, 141, 143, 144, 146, 159\)[/tex]
Since there are 10 numbers (an even number), the median will be the average of the 5th and 6th numbers in this ordered list:
[tex]\[
\text{Median} = \frac{141 + 141}{2} = 141.0
\][/tex]
3. Mode:
The mode is the number that appears most frequently in the list.
From the list [tex]\(141, 144, 133, 128, 141, 136, 146, 159, 143, 141\)[/tex], the number [tex]\(141\)[/tex] appears most frequently (3 times).
Mode:
[tex]\[
\text{Mode} = 141
\][/tex]
Summary of Question 22:
- Mean: [tex]\(141.2\)[/tex]
- Median: [tex]\(141.0\)[/tex]
- Mode: [tex]\(141\)[/tex]
Question 23:
Evaluate if [tex]\(x = 7\)[/tex], [tex]\(y = 4\)[/tex], and [tex]\(z = 4\)[/tex]:
[tex]\[
\left(y^2 - x^2\right) \div (9z - 2)
\][/tex]
Substitute the given values into the expression:
[tex]\[
x = 7, \quad y = 4, \quad z = 4
\][/tex]
First, calculate [tex]\(y^2 - x^2\)[/tex]:
[tex]\[
y^2 = 4^2 = 16
\][/tex]
[tex]\[
x^2 = 7^2 = 49
\][/tex]
[tex]\[
y^2 - x^2 = 16 - 49 = -33
\][/tex]
Next, calculate [tex]\(9z - 2\)[/tex]:
[tex]\[
9z = 9 \cdot 4 = 36
\][/tex]
[tex]\[
9z - 2 = 36 - 2 = 34
\][/tex]
Finally, divide [tex]\((y^2 - x^2)\)[/tex] by [tex]\((9z - 2)\)[/tex]:
[tex]\[
\frac{y^2 - x^2}{9z - 2} = \frac{-33}{34} \approx -0.9705882352941176
\][/tex]
Summary of Question 23:
The value of the expression is approximately [tex]\(-0.9705882352941176\)[/tex].
I hope this helps! If you have any other questions, feel free to ask.
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