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podryga [215]
4 years ago
6

How many vertices does a square pyramid have

Mathematics
1 answer:
Afina-wow [57]4 years ago
5 0

Answer:

5 vertices

Step-by-step explanation:

4 on the bottom and 1 on the top

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The data set represents the number of rings each person in a room is wearing. 0, 2, 4, 0, 2, 3, 2, 8, 6 What is the interquartil
forsale [732]

Answer:

IQR=4

Step-by-step explanation:

We have been given a data set that represents the number of rings each person in a room is wearing as: 0, 2, 4, 0, 2, 3, 2, 8, 6. We are asked to find the interquartile range of the data.

Let us arrange our data points from least to greatest order.

0, 0, 2, 2, 2, 3, 4, 6, 8.

Let us find median of our data set. Since our data set has 9 data points, therefore, median of our given data set will be value of 5th term, that is 2.

Now let us find lower quartile, which is also known as the median of the data points to the left of median.

We have 4 data points to the left of our median, so our lower quartile will be average of 2nd and 3rd data point.

Q_1=\frac{0+2}{2}=\frac{2}{2}=1

Now let us find upper quartile of our given data set, which is the median of the data points to the right of median.

We have 4 data points to the right of the median, so our upper quartile will be the average of 7th and 8th data point.

Q_3=\frac{4+6}{2}=\frac{10}{2}=5

Since we know that interquartile range is the difference between upper and lower quartile.

IQR=Q_3-Q_1

IQR=5-1

IQR=4

Therefore, IQR of our given data set is 4.

8 0
3 years ago
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What is the value of x?<br><br><br><br> Enter your answer in the box.<br><br> x =
Oliga [24]

\star \blue{ \frak{To \: find :}}

\\  \\

  • value of x

\\  \\

\star \blue{ \frak{solution:}}

\\  \\

So to find value of x , we have to apply Linear Pair.

\\  \\

<u>Equation formed:</u>

\\  \\

\bigstar \boxed{ \tt(10x - 20) \degree + (6x + 8)\degree = 180 \degree}  \\

\\  \\

<u>Step by step expansion:</u>

\\  \\

\dashrightarrow  \sf(10x - 20) \degree + (6x + 8)\degree = 180 \degree \\

\\  \\

\dashrightarrow  \sf10x - 20 \degree + 6x + 8\degree = 180 \degree \\

\\  \\

\dashrightarrow  \sf10x +6x- 20 \degree  + 8\degree = 180 \degree \\

\\  \\

\dashrightarrow  \sf16x- 20 \degree  + 8\degree = 180 \degree \\

\\  \\

\dashrightarrow  \sf16x- 12\degree = 180 \degree \\

\\  \\

\dashrightarrow  \sf16x = 180 \degree + 12\degree\\

\\  \\

\dashrightarrow  \sf16x =192\degree\\

\\  \\

\dashrightarrow  \sf \: x = \frac{192\degree}{16\degree} \\

\\  \\

\dashrightarrow  \sf \: x = 12  \degree

\\  \\

\therefore \underline {\textsf{\textbf{Value of x is \red{12\degree}}}}

4 0
2 years ago
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