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faust18 [17]
3 years ago
10

A box-and-whisker plot. The number line goes from 0 to 9. The whiskers range from 0 to 7, indicated by A and E. The box ranges f

rom 2 to 6, indicated by B and D. The line divides the box at 4, indicated by C.
Identify the location of the values used to create a box plot.

A =
✔ minimum value

B =
✔ lower quartile

C =
✔ median

D =
✔ upper quartile

E =
✔ maximum value
Mathematics
1 answer:
Sladkaya [172]3 years ago
6 0

minimum value: 1

lower quartile: 3

median of the data: 6

upper quartile: 8

maximum value: 14

Sorry for the late answer, i hope this will help you in the future and everyone else who reads this.

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Round the number 8.946 to the nearest tenth
Mumz [18]

Answer:

8.9

Step-by-step explanation:

Because it is a 4, round down.

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4 years ago
Write the following equation in slope intercept form.<br><br> 7x-4y=22<br><br> y= x+
Finger [1]

Answer:

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Step-by-step explanation:

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What is the slope of the line<br>​
charle [14.2K]

Answer:

The slope is -1/1 or just -1.

(When you need to plot the slope, you use -1/1 to use the strategy Rise/Run or Rise over Run.

If you look at the point (-3,1) From that point, you go down -1 and you run to the side for a positive 1.

Once you do that you'll land at (-2,0) and you can keep on repeating this and you'll see that it lands on the points on the line.

Hope this helps! Please feel free to ask any questions if needed! :)

4 0
3 years ago
T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


{y}''=7e^{x}\\ {y}'=7e^{x}


k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

7(e^x)(1+49e^{2x})(98e^{2x}-1)=0

→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

e^{x}=0  ,  ∧ 1+49e^{2x}=0   [not possible ∵there exists no value of x satisfying these equation]

→98e^{2x}-1=0

Solving this we get

x= -\frac{1}{2}\ln{98}

As you will evaluate {k(x})}''<0 at x=-\frac{1}{2}\ln98

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[-\frac{1}{2}\ln98,1/√2]

k(x)=\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}

k(x)=\frac{7}{\infty}

k(x)=0







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Step-by-step explanation:

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