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3 years ago

Which values in this data set are outliers?

Mathematics

2 answers:

borishaifa [10]3 years ago

5 0

21 is your outlier becuase it is the farthest away from everyone else

stellarik [79]3 years ago

4 0

The number that is the farthest away from every other number. The answer is 21.<span />

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The equation y — 480 = – 120 (x - 1) represents the total number of kilometers, y, ben and his family have left to drive to Ben'

MAXImum [283]

Answer:

First answer is 4 and the next is 120

7 0

3 years ago

Determine whether the vectors u and v are parallel, orthogonal, or neither. u = <6, -2>, v = <8, 24>

Leno4ka [110]

Because 6 * 8 + ( - 2 )* 24 = 48 - 48 = 0 , the vectora u and v are orthogonal ;
Two vectors are orthogonal if their dot product is zero.<span />

5 0

3 years ago

HELPPP I NEED HELP DUE VERY SOONNN

Dovator [93]

Answer:

x . 1 . 3 . 5 . 7 . 9

y . 6 . 18 . 30 . 42 . 54

Step-by-step explanation:

X-values are the odd numbers

Y-values are the corresponding x-values multiplied by 6

8 0

3 years ago

A sporting goods store sells two types of snowboards. Brand S sells for $309. Brand B sells for $489. Last month, the store sold

avanturin [10]

Let S be the number of Brand S snowboards sold.
Let B be the number of Brand B snowboards sold.

309S + 489B = 4299
S + B = 11

S = 6 snowboards sold
B = 5 snowboards sold

6 0

3 years ago

Prove that a triangle with the sides a - 1 cm to root under a

vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm

Hence,

${AB}^2 = (a -1)^2$

Now We know that

$(a- b)^2 = a^2+b^2 - 2ab$

So;

${AB}^2= a^2 + 1^2 -2\times a \times1$

${AB}^2 = a^2 + 1 -2a$

Now;

${BC}^2 = (2\sqrt{a})^2= 4a$

Also;

${CA}^2 = (a + 1)^2$

Now We know that

$(a+ b)^2 = a^2+b^2 + 2ab$

${CA}^2= a^2 + 1^2 +2\times a \times1$

${CA}^2 = a^2 + 1 +2a$

${CA}^2 = AB^2 + BC^2$

[By Pythagoras theorem]

$a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a$

Hence, ${CA}^2 = AB^2 + BC^2$

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0

3 years ago