What is the median of 79,85,86,90,92,93

Mathematics

1 answer:

o-na [289]2 years ago

7 0

THE MEDIAN IS THE MIDDLE NUMBER IS 86 AND 90

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FIND 2 EQUIVALENT FORMS FOR EACH FRACTION USING MULTIPLICATION AND DIVISION. 8/12,6/8,9/15,2/16

Alja [10]

8/12: 2/3 and 16/24
6/8: 3/4 and 12/16
9/15: 3/5 and 18/30
2/16: 1/8 and 4/32

5 0

3 years ago

Read 2 more answers

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

2 years ago

What are the x-intercept and y-intercept of the line passing through (0,-3) and (5,0)

Butoxors [25]

Answer:

y-intercept: (0, -3)

x-intercept: (5, 0)

Step-by-step explanation:

Trick question! The problem gives you the x and y intercepts. x intercept is when y is 0, and y intercept is when x = 0!

3 0

3 years ago

HELP

Lostsunrise [7]

Answer:

${b}^{2} + 3b - 3 = 0$

Step-by-step explanation:

${b}^{2} + 3b - 3 = 0 \\ because \: \: {b}^{2} < 4ac \\ {b}^{2} = {3}^{2} = 9 \\ 4ac = 4 \times 1 \times - 3 = - 12 \\ hence : {b}^{2} < 4ac$

3 0

2 years ago

R=8 P=7 R+P= PLZ im stuck <br>

mihalych1998 [28]

Answer:

Step-by-step explanation:

The equation is just a plug in and solve. you have your given variables and all you have to do it plug them into the matching spot and solve

R=8 P=7

8+7=15

7 0

3 years ago