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3 years ago

Given the following set of data, which measures have a value of 6? (Check all that apply.)

Mathematics

2 answers:

scoray [572]3 years ago

7 0

Range is subtracting the biggest value and the smallest value of the data. For this set of data the range is:

9 - 3 -------------------------------> 6

Mode is the number that appears most often. In this set of data no number appears more than twice so mode is not applicable

Median is the middle of the numbers when ordered from least to greatest. For this set of the data the median is:

3 4 6 8 9

4 6 8

Mean is adding all the number together then dividing the sum by how many numbers there are in the data. For this data the mean is:

(3 + 4 + 6 + 8 + 9) ÷ 5

(30) ÷ 5 ------------------------------------> 6

As you can see the range, median, and mean all have the value of 6

Hope this helped!

morpeh [17]3 years ago

7 0

Answer:

titi's

Step-by-step explanation:

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What is the answer to this

deff fn [24]

So the first column is if a serving is 36 so all you have to do is divide all the different ingredient amounts by three and that will tell you the answer for the serving of 12 row because 12 * 3 = 36

and for the 24 row just multiply all the numbers you get in the 12 row and that will be the answer cause 12 * 2 = 24

then what ever numbers you get on the 12 serving column just divide those by 12 and that will tell you how much you get for one serving and that will help you solve for the 60 one and the 300 one all you have to do it times the numbers by 60 and then 300

8 0

3 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

8 0

2 years ago

HELP ME PLEASE PLEASE IM BEGGING

suter [353]

Answer:

FALSE, (2, 9) is not a solution to the set of inequalities given.

Step-by-step explanation:

Simply replace x by 2 and y by 9 in the inequalities and see if the inequality is true or not:

irst inequality:

$y\geq 4x\\9\geq 4\,(2)\\9\geq 8$

so thi inequality is verified as true since 9 is larger or equal than 8

Now the second inequality:

$y$

This is FALSE since 9 is larger than 4 (not smaller)

Therefore the answer to the question is FALSE, (2, 9) is not a solution to the set of inequalities given.

6 0

3 years ago

PLEASE ANSWERRRRRRRRR

chubhunter [2.5K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Need help with all And have to show work

kari74 [83]

Altho' I can easily guess what you're supposed to do here, I must point out that you haven't included the instructions for this problem.

I'll help you by example. Let's look at the first problem:

"Evaluate 6(z-1) at z-4."

Due to "order of operations" rules, we must do the work inside the parentheses FIRST. Replace the z inside (z-1) with "-4". We obtain

6(-4-1) = 6(-5) = -30 (answer.)

Your turn. Try the next one. If it's unclear, as questions.

8 0

3 years ago