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Phoenix [80]
2 years ago
7

It is best to use median as your measure of center when you have an extreme number in your set of data (also known as an outlier

).
Mathematics
1 answer:
qwelly [4]2 years ago
5 0

Answer:

true

Step-by-step explanation:

The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center. The words “mean” and “average” are often used interchangeably.

You might be interested in
Single quadrant ordered pairs
Marianna [84]

Answer:

this four sections are called the quadrants

3 0
3 years ago
How do I solve this, and what is the answer? please and thank you.
vovangra [49]

Answer:

A. (x +8) + (-4x+31)/(x^2+2x+1)

Step-by-step explanation:

When you perform long division of polynomials, the first quotient term is the ratio of the highest-degree terms in the numerator and denominator: x^3/x^2 = x.

This fact eliminates all but choices A and C.

The denominator of the remainder term is the denominator of the original expression, so will be x^2 +2x +1, as shown in choice A.

__

So, simply based on a couple of facts about long division (that you learned in the early elementary grades), you can make the correct choice of answer without actually working the problem in detail.

_____

This is a polynomial long division problem. It is worked in virtually the same way that numerical long division problems are worked: first you find a quotient term, then you multiply that by the divisor and subtract the result from the dividend. The difference is the new dividend. These steps are identical to numerical long division.

For polynomial long division, instead of lining up the digits with the same place value, you line up the terms with the same degree of the variable.

As mentioned above, the quotient term is computed only from the highest-degree terms of dividend and divisor, so that part is actually simpler than for numerical long division.

A dividend that is of lower degree than the divisor is considered to be the remainder. As with numerical long division, it can be expressed as a fraction with the divisor as the denominator.

Numerical example: 18/7 = 2 remainder 4 or 2 4/7.

8 0
3 years ago
68 =34 =23<br><br> True or False? (Hint: Type true or false)
frosja888 [35]

Answer:

I believe it's false

Step-by-step explanation:

3 0
2 years ago
Ryan works at at the donut shop where he makes $10.25 per hour . He also works part time at the school bookstore where he makes
SCORPION-xisa [38]

Answer:

6 hours

Step-by-step explanation:

Let

x----> the number of hours worked at the donut shop

y----> the number of hours worked at the school bookstore

we know that

x+y=20

x=20-y -----> equation A

10.25x+8.75y=196 -----> equation B

Substitute equation A in equation B and solve for y

10.25(20-y)+8.75y=196

205-10.25y+8.75y=196

10.25y-8.75y=205-196

1.5y=9

y=6 hours at the school bookstore

7 0
3 years ago
Solve x^3-7x^2+7x+15​
ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0
3 years ago
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