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iVinArrow [24]
3 years ago
5

The correct answer to the question *interquartile range

Mathematics
2 answers:
Vanyuwa [196]3 years ago
7 0
The interquartile range (IQR) is 20

You find this by subtracting the values of Q3 and Q1
Q3 is the right most edge of the box which is 45
Q1 is the left most edge of the box which is 25

IQR = Q3 - Q1 = 45 - 25 = 20

Side Note: The median is not 45. The median is actually 40 since this is the middle line of the box, which is in between 35 and 45
mars1129 [50]3 years ago
5 0
The correct answer is 20
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Please need help in these 3 algebra questions !!!!
Marysya12 [62]

Answer:

\large\boxed{7.\ B.\ 8s^2+16s+1}\\\\\boxed{8.\ D.\ -15t^9u^{11}}\\\\\boxed{11.\ C.\ 3,\ -2}

Step-by-step explanation:

7.\\(3s^2+7s+2)+(5s^2+9s-1)=3s^2+7s+2+5s^2+9s-1\\\\\text{combine like terms}\\\\=(3s^2+5s^2)+(7s+9s)+(2-1)\\\\=8s^2+16s+1

8.\\(-3t^2u^3)(5t^7u^8)=(-3\cdot5)(t^2t^7)(u^3u^8)\qquad\text{use}\ a^na^m=a^{n+m}\\\\=-15t^{2+7}u^{3+8}=-15t^9u^{11}

11.\\n-the\ number\\\\n^2=n+6\qquad\text{subtract}\ n\ \text{and}\ 6\ \text{from both sides}\\\\n^2-n-6=0\\\\n^2+2n-3n-6=0\\\\n(n+2)-3(n+2)=0\\\\(n+2)(n-3)=0\iff n+2=0\ \vee\ n-3=0\\\\n+2=0\qquad\text{subtract 2 from both sides}\\n=-2\\\\n-3=0\qquad\text{add 3 to both sides}\\n=3

4 0
3 years ago
A king size candy bars costs $1 with each candy bar having 1,472 calories. If you bought 7 candy bars and took 8 days eating the
Gre4nikov [31]

Answer: your mom gay ;)

Step-by-step explanation:

4,500 calories a day

If you ate the same amount each day, that means you would eat 3 a day, since 6 divided by 2 is 3.

Find how many calories you would eat each day by multiplying 3 by 1,500

3(1,500)

= 4500

5 0
3 years ago
Find the distance between the pair of points: (6,−3) and (1,0
rosijanka [135]

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

\sqrt{34}

»»————- ★ ————-««  

Here’s why:

⸻⸻⸻⸻

\text{\underline{The distance formula is:}}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

⸻⸻⸻⸻

\boxed{\text{Calculating the answer....}}\\\\d = \sqrt{(1-6)^2+(0-(-3))^2}\\--------------\\\rightarrow d = \sqrt{(-5)^2+(3)^2}\\\\\rightarrow d = \sqrt{(25+9)}\\\\\rightarrow d =\boxed{ \sqrt{34} }

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.

6 0
3 years ago
Statistics when mode is best measure of central tendency.
ehidna [41]

Answer:

The mode is the least used of the measures of central tendency

Step-by-step explanation:

6 0
2 years ago
find the slope of the curve y=x^2-2x-5 at the point P(2,5) by finding the limit of secant slopes through point P
Fynjy0 [20]

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, (t,y(t))=(t,t^2-2t-5), where t>2. The slope of the secant line through P and Q is given by the difference quotient,

\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t

where we are allowed to simplify because t\neq2.

Then the equation of the secant line is

y-(-5)=t(x-2)\implies y=t(x-2)-5

Taking the limit as t\to2, we have

\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

\dfrac{\mathrm dy}{\mathrm dx}=2x-2

and at x=2, we get a slope of 2(2)-2=2, as expected.

4 0
3 years ago
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