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Ad libitum [116K]
3 years ago
7

How to find the new interquartile range of the numbers 9,10,11,12,16,20,20

Mathematics
2 answers:
UNO [17]3 years ago
6 0

10 is ur range

hope this helps

muminat3 years ago
6 0

Answer:

The interquartile range would be 10.

Step-by-step explanation:

9, 10, 11, 12, 16, 20, 20

Median: 12

Lower quartile (Q1): 10

Upper quartie (Q2): 20

(IQR) Interquartile range: 20 - 10 = 10

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I have 1 litre of water in a jug. On the way to my table I spill some. I have 780ml left. How much have I lost?​
irina1246 [14]

Step-by-step explanation:

1 litre = 1000ml

780ml left

lost amount = 1000 - 780 = 220ml

3 0
2 years ago
If triangle XYZ is reflected across the line y = 1 to create triangle X'Y'Z', what is the ordered pair of X'? 
Vlad1618 [11]

Consider triangle XYZ with vertices at points X(1,-3), Y(3,0) and Z(-1,-1). If triangle XYZ is reflected across the line y = 1, then the rule of reflection is

(x,y)→(x,-y+2).

The image X' of point X will have coordinates according to the given rule:

X(1,-3)→X'(1, -(-3)+2)=X'(1,5).

Answer: correct choice is D.

3 0
3 years ago
Read 2 more answers
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
11 months ago
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dimaraw [331]
Can you show us the triangular prism to help you
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A publisher shipped 15 boxes of books to a bookstore each box contained 32 books how many books did the publisher ship to the bo
Alenkinab [10]

Answer:

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Step-by-step explanation:

8 0
2 years ago
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