How can data range be calculated?a.
Differences between mean and mode

b.
Differences between mean and median

c.
Differences between median and mode

d.
Differences between max value and min value

Answers

Answer 1
Answer:

Answer:

Option: d is the correct answer.

  d.    Differences between maximum value and minimum value.

Step-by-step explanation:

We know that the data range i.e. Range of the data is the difference between the largest (i.e. maximum value) data value and the smallest (i.e. minimum value) data value.

It is calculated in following steps--

  • We arrange our data in the increasing or ascending order.
  • Then we find the difference between the maximum and the minimum value of the data set.


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Answers

Answer:

  24 crates

Step-by-step explanation:

The usable load capacity for 40-kg crates is ...

  1050 kg - 82 kg = 968 kg

The number of 40-kg crates that can be loaded in a trailer with this capacity is ...

  floor(968/40) = floor(24.2) = 24 . . . . crates

_____

The function floor(n) gives the greatest integer less than or equal to n. For positive numbers, it drops the fraction.

Subject: writing equations
Fake answers-reported cuz I’m rly tryna pass this shi.

Answers

Answer:

1) Y-6=-3/2(X-0)     2) Y= -3/2x+6   3) 3x+2y=12

Step-by-step explanation:

Q4, PLSS HELP, IT TIME.........

Answers

Answer:

dilation:) dilation always produces a congruent figure...

8 5/12 divided by 1 3/4

Answers

^fractions lesson^

8 5/12 : 1 3/4

step 1:

(8 × 12 + 5)/12

=101/12

step 2:

(1 × 4 + 3)/4

=7/4

ok, enter:

101/12 : 7/4

=101/12 × 4/7

=404/84

=4 68/84

=4 17/21

Hello can you please help me posted picture of question

Answers

This is true statement.

Sample space is made from all the possible outcomes that an event can have.
For example when tossing a coin, the possible outcomes are Head and Tail so the sample space will be {Head, Tail}.

Thus, option A is the correct answer

How did you do this question?

Answers

Answer:

R = ∞

I = (-∞, ∞)

Step-by-step explanation:

Use the ratio test:

lim(n→∞)│aₙ₊₁ / aₙ│

lim(n→∞)│[xⁿ⁺⁶ / (2(n+1)!)] / [xⁿ⁺⁵ / (2n!]│

lim(n→∞)│[xⁿ⁺⁶ / (2(n+1)!)] × (2n! / xⁿ⁺⁵)│

lim(n→∞)│x 2n! / (2(n+1)!)│

lim(n→∞)│n! / (n+1)!││x│

lim(n→∞) (1 / (n+1))│x│

0

The series converges if the limit is less than 1.

The limit is always less than 1, so the radius of convergence is infinite.

So the interval of convergence is (-∞, ∞).