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avanturin [10]
3 years ago
10

the question is on the photo everything you guys need to know is there pls help me or I won't be able to turn in my homework.Thx

to all the people that helped me ( well the people that answered )

Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

The first and third quartiles were calculated incorrectly. Also, this means the inter quartile range is incorrect.

Step-by-step explanation:

The first quartile is going to be the median of the lower half of the data set {89, 93, 99, 110}  this median is (93+99)/2, or 96.

The third quartile can be calculated the same way with the data set {135, 144, 152, 159} This gives us 148 as the third quartile.

Lastly, the interquartile range is just the difference between the first and third quartiles, which is 52.

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Round each number to the nearest tenth.
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Step-by-step explanation: Any number that is 5 or greater can be used to round any number.


7 0
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Solve for x: |2x + 12| = 18
julsineya [31]
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8 0
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Multiply 3 by the number of days in a week. Subtract 12 and write your
k0ka [10]

Answer:

9?

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nlexa [21]
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3 years ago
Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.
andrey2020 [161]
Check the picture below.

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keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}})
\\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------

\bf \begin{cases}
h=4\\
k=1\\
p=4
\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)
\\\\\\
\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y

8 0
3 years ago
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