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3 years ago

A _________ is a value that describes the center of a data set and includes the mean, median, and mode.

2 answers:

7 0

A central tendency is a value that describes the center of a set and includes the mean, median, and mode.

Mean is Something which is intermediate or in the middle; an intermediate value or range of values; a medium..

Median The quantity or value at the midpoint of a set of values, such that the variable is equally likely to fall above or below it; the middle value of a discrete series arranged in magnitude (or the mean of the middle two terms when there is an even number of terms).

Mode is the most frequently occurring value in a distribution

ladessa [460]3 years ago

4 0

A central tendency is a value that describes the center of a data set and includes the mean, median, and mode.

And just in case you don't know:

Mean is the sum of all the numbers divided by how many numbers there are. Basically the average.

Mode is the number that occurs the most times.

Median is the middle number when you write the numbers from lowest to greatest.

You might be interested in

How can we classify the following shapes ? Please choose correct answers that apply !!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!

liq [111]

Answer:

Step-by-step explanation:

Cause a square has all equal sides

5 0

3 years ago

Drag each symbol and number to the correct location on the inequality. Not all symbols and numbers will be used. Sam initially i

galben [10]

Answer:

$t \leq 18.67\ years$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=?\ years\\ P=\$4,500\\ A=\$7,020\\r=0.03$

The inequality that represented this situation is

$P(1+rt) \leq A$

substitute the values and solve for t

$4,500(1+0.03t) \leq 7,020$

$(0.03t) \leq (7,020/4,500)-1$

$t \leq [(7,020/4,500)-1]/0.03$

$t \leq 18.67\ years$

7 0

3 years ago

State the rule obeyed by the variables in each relation diagram below

labwork [276]

Answer:

y = 2x + 1

Step-by-step explanation:

You can see a pattern in the group of x's that the numbers go up by 1 Also in the y-set that the numbers go up by 2. So this pattern is linear, that means the "rule" you are looking for does not have exponents or square roots or any very complicated stuff. You can use a guess and check method. Say to yourself how can I get a 7 out, when I put a 3 in? "times by2 and plus 1" works.

3 times 2, and plus1

gives you 7.

Test it on the other numbers.

-1 times2, and plus1

2(-1)+1 = -1

2(0)+1 = 1

2(1)+1 = 3

2(2)+1 = 5

It works for all the numbers.

You can calculate it also, using any two pairs of (x,y) from the data set. Put y-y on top of a fraction and x-x on the bottom. You will get the slope and that is the 2 in the "rule"

(3,7) and (2,5) for example. 7-5 so put 2 on top and 3-2 so put 1 on the bottom. 2/1 is just 2. From the point (0,1) we know the y-intercept is 1. This also gives the equation y=2x+1.

If you are just starting to learn this, probably just guess and check a rule. The rule has to work for all the points.

6 0

2 years ago

2(c+1)=10 HELP PLSSSSSSSS

Nikolay [14]

Answer:

c=4

Will venus1234 delete the c4?

Step-by-step explanation:

You simplify by distributive property

2c+2=10

Then you can use subtraction property of equality to subtract 2 from both sides to get

2c=8

Then do division property of equality to get the unit rate of c.

You end up with

c=4

Hope this helps!

5 0

3 years ago

Read 2 more answers

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$