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Paul [167]
3 years ago
6

Find interquartile range for 3,4,5,6,7

Mathematics
2 answers:
lions [1.4K]3 years ago
8 0
3, 4, 5, 6 and 7
Q 1= (3+4)/2=3.5
Q 2= (4+5)/2=4.5
Q 3=(5+6)/2=5.5
Interquartile range =Q3-Q1
=5.5-3.5
=2
worty [1.4K]3 years ago
8 0

Median: 5

Lower Quartile: 3.5

Upper Quartile: 6.5

IQR Interquartile Range: 6.5 - 3.5 = 2

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Find the fifth roots of 243(cos 260° + i sin 260°)
Charra [1.4K]
<span>use De Moivre's Theorem:

⁵√[243(cos 260° + i sin 260°)] = [243(cos 260° + i sin 260°)]^(1/5)

= 243^(1/5) (cos (260 / 5)° + i sin (260 / 5)°)

= 3 (cos 52° + i sin 52°)

z1 = 3 (cos 52° + i sin 52°) ←← so that's the first root


there are 5 roots so the angle between each root is 360/5 = 72°

then the other four roots are:

z2 = 3 (cos (52 + 72)° + i sin (52+ 72)°) = 3 (cos 124° + i sin 124°)

z3 = 3 (cos (124 + 72)° + i sin (124 + 72)°) = 3 (cos 196° + i sin 196°)

z4 = 3 (cos (196 + 72)² + i sin (196 + 72)°) = 3 (cos 268° + i sin 268°)

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3 0
3 years ago
I need help pls!! 20 points I need it!!
dolphi86 [110]

Answer: About 6.71 units

Step-by-step explanation:

You use the distance formula: \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}  }.

Assign the the two points to the variables:

(x_{1}, y_{1}) =  (1, 3)\\(x_{2}, y_{2}) = (4, -3)

Substitute them in the formula:

\sqrt{(4-1)^{2}+(-3-3)^{2}}

And solve it:

\sqrt{(4-1)^{2}+(-3-3)^{2}}\\=\sqrt{(3)^{2}+(-6)^{2}}\\=\sqrt{9+36}\\=\sqrt{45}\\=6.71

3 0
3 years ago
Read 2 more answers
F(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-%5Csqrt%7Bx%5E2%2B2x-15%7D" id="TexFormula1" title="\sqrt{x+7} -\
elena-s [515]

Answer:

x >= -7  ................(1a)

x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

x+7 >= 0  .....................(1)

AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

(x+5)(x-3) >= 0

This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

OR

x <= -5 ................(2b1)

)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

3 0
3 years ago
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The longest side of an obtuse triangle measures 20 cm. The two shorter sides measure x cm and 3x cm.
laiz [17]
The answer to your question would be 6.3
7 0
3 years ago
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Answer: what are you trying to ask here?

Step-by-step explanation:

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