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pantera1 [17]
3 years ago
15

What is the mean, median, interquartile range, and standard deviation of 20, 20, 28, 28, 30, 30, 30, 36, 36, 40, 40

Mathematics
1 answer:
Kazeer [188]3 years ago
4 0
Median is 30 range is 20 and that's all I'm gonna do sorry I have my own math hope I helped
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Please help ASAP <br> I really need help please will name you the brain
MAXImum [283]

Answer:

I believe answer B

Step-by-step explanation:

It says shifted UP 3 units and 2+3=5 so that is my answer. Feel free to yell at me if I am wrong.

8 0
3 years ago
Find two consecutive odd integers whose product is 323
Harrizon [31]
Use a let statement 
first
let x and x + 2 be the number so you write it like this
<u>let x = the first consecutive integer
</u><u>let x + 2 = the second consecutive integer
</u>
second
x(x+2)=323
x^2 + 2x = 323
        -323  -323
x^2 + 2x -323 = 0

third
try to factor -323 so it is 19 and -17
(x + 19) (x - 17) = 0
x = 19 
x = -17

hope this help

4 0
3 years ago
Plz help me w this !!
larisa [96]

Answer:

sum of angles in atriangle=180°

Step-by-step explanation:

w-45°+w-37°+w+19°=180°(sum of angles in a triangle)

3w=180°+45°+37°-19°

3w=243°

w=81°

7 0
3 years ago
2x + 16 = 22x<br><br> Variables on both sides
Nonamiya [84]

answer: Subtract sixteen from both sides of the equation

2 +16 -16=22x -16

x = 4/5

8 0
2 years ago
The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
3 years ago
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