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4 years ago

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Which statements are true about the median of a data set? Check all that apply. The median isn’t affected much by one outlier. T

he median is always affected by one outlier. The median is the number in the middle on an ordered set of data. To find the median, find the difference between the highest and lowest numbers. To find the median of an even data set, find the average of the two middle numbers.

2 answers:

sasho [114]4 years ago

7 0

These are all the right answers:

The median isn’t affected much by one outlier.

The median is the number in the middle of an ordered set of data.

To find the median of an even data set, find the average of the two middle numbers.

Hope this helped :)

Arturiano [62]4 years ago

5 0

To find the median of an even data set, find the average of the two middle numbers.

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Please answer correctly and no links

victus00 [196]

Answer:

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Step-by-step explanation:

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3 years ago

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

4 years ago

I’m confused please solve it.

Allushta [10]

Answer:

b = 65

Step-by-step explanation:

b is the y- intercept

if you start to graph 6x from the two points that are already given then it'll find the y-intercept which is b

here's the graph

3 0

3 years ago

Can somebody help with this problem?

Anastasy [175]

Answer:

x=-5 or -5

Step-by-step explanation:

x - 4 = 2x + 1

Subtract x from both sides canceling out the x on the left, getting...

-4 = x + 1

Then subtract the 1 from both sides canceling out the 1 on the right, getting...

-5 = x

Subtracting x from 2x gets you 1x which you just write as x because there is only one x. Doing the equation from the other side will still get you the same answer and I will show you.

x - 4 = 2x + 1

-x - 4 = 1

-x = 5

x = -5

Hope this helps!

4 0

3 years ago

Joel scores in 80% of his games. His team is starting the playoffs, and if they make it to the state championship, they will hav

Fittoniya [83]

Answer:

D

Step-by-step explanation:

4 out of 5 games is 80%:

So if we know, as a rule that "Joel scores in 80% of his games.", the probability that he will score in 80% of his games is 1 - that is, we know it for sure

100% sure

4 0

3 years ago

Read 2 more answers