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Alika [10]
3 years ago
5

I have many students are represented in the data set?

Mathematics
1 answer:
nalin [4]3 years ago
7 0

Answer:

a) 19 students

b) the mean 3 9/19

the median 3

the mode 3

c) the range 6

Step-by-step explanation:

The data set shows that

0 points gets 1 student,

1 point gets 1 student,

2 points get 2 students,

3 points get 6 students,

4 points get 4 students,

5 points get 3 students and

6 points get 2 students.

a) There are 1+1+2+6+4+3+2=19 students.

b) The mean is

\dfrac{1\cdot 0+1\cdot 1+2\cdot 2+6\cdot 3+4\cdot 4+3\cdot 5+2\cdot 6}{19}=\dfrac{66}{19}=3\dfrac{9}{19}.

The average score is 3 9/19 points.

The median is 10th term in the data set - 3 points (means the middle score in the data set)

The mode is 3 points (means most happened score)

c) The range of the data is 6-0=6 points.

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A rectangle is 4xcm wide and 2cm long what do you notice
deff fn [24]

The width of the rectangle is twice the length of the rectangle

<h3>How to compare the dimensions of the rectangle?</h3>

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4 0
1 year ago
Use Synthetic Division to divide 6x^3-10x^2+20 by x+1<br><br> Quotient:_____<br><br> Remainder:_____
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Answer:

the quotient is 6x^2 - 16x + 16, and the remainder is just 4

Step-by-step explanation:

The polynomial 6x^3-10x^2+20 has coefficients {6, -10, 0, 20}.  Division by the binomial x + 1 requires that we use -1 as the divisor.  The synthetic division setup becomes:

-1   /    6    -10    0    20

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     -----------------------------

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6 0
3 years ago
What is the vale of ■ in this equation?<br>7×9=(7×10)-(7×■)
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5 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%20%5Csf%20%5Chuge%7B%20question%20%5Chookleftarrow%7D" id="TexFormula1" title=" \sf \huge
BabaBlast [244]

\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

\underline{\large{\bf Identities\:used:-}}

\boxed{\sf (a+b)^2=a^2+2ab+b^2}

\boxed{\sf (√a)^2=a}

\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}

\boxed{\sf \sqrt{\sqrt{a}}=a}

\underline{\bf Final\: Solution:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}

\bull\sf Apply\: Squares

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

\bull\sf Put\:values

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}

\underline{\bf More\: simplification:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

\underline{\Large{\bf Simplified\: Answer:-}}

\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}

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