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gavmur [86]
3 years ago
9

A/4 − 5/6 = −1/2. Solve the equation.

Mathematics
2 answers:
Wewaii [24]3 years ago
8 0

Answer:

=-12

Step-by-step explanation:

julsineya [31]3 years ago
4 0

Answer:

a = 4/3

Step-by-step explanation:

hmxgnxcgnxgnxyjxkydyjdyjd

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The following data show the scores of some students in a competition:
Dima020 [189]

The box plot that represents the data is a box plot titled "Scores of Participants" and labeled "Score" uses a number line from 10 to 35 with primary markings and labels at 10, 15, 20, 25, 30, and 35. The box extends from 13 to 27 on the number line. A line in the box is at 24. The whiskers end at 11 and 31.(second option)

<h3>Which box plot represents the data?</h3>

A box plot is used to study the distribution and level of a set of scores. The whiskers represent the minimum and maximum values.

On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median. The third line on the box represents the upper (third) quartile. 75% of the scores represents the upper quartile.

The data arranged in ascending order : 11, 13, 23, 24, 24, 27, 31

Median = 24

First quartile = 1/4 x (7 + 1) = 2nd term =13

Third quartile = 3/4 x (7 + 1) = 6th term = 27

To learn more about box plots, please check: brainly.com/question/27215146

#SPJ1

6 0
2 years ago
If you draw four cards at random from a standard deck of 52 cards, what is the probability that all 4 cards have distinct charac
mezya [45]

There are \binom{52}4 ways of drawing a 4-card hand, where

\dbinom nk = \dfrac{n!}{k!(n-k)!}

is the so-called binomial coefficient.

There are 13 different card values, of which we want the hand to represent 4 values, so there are \binom{13}4 ways of meeting this requirement.

For each card value, there are 4 choices of suit, of which we only pick 1, so there are \binom41 ways of picking a card of any given value. We draw 4 cards from the deck, so there are \binom41^4 possible hands in which each card has a different value.

Then there are \binom{13}4 \binom41^4 total hands in which all 4 cards have distinct values, and the probability of drawing such a hand is

\dfrac{\dbinom{13}4 \dbinom41^4}{\dbinom{52}4} = \boxed{\dfrac{2816}{4165}} \approx 0.6761

4 0
2 years ago
6(840.25)+15(1,374.05)+9(1.222.43)/30
kifflom [539]

Answer:

26018.979

yea...

4 0
3 years ago
Which ordered pair is a solution of the equation?
GrogVix [38]

Answer

A only (2,3)

Step-by-step explanation:

If you graph it or plug it into Desmos that can also help

6 0
3 years ago
the coordinates of the endpoints of BC are B(3,2) and C(8,17). Point D is on BC and divides it such that BD:CD is 3:2. What are
eduard
Snsjdnsndnxnendndnfbdnxndjdnxjxhsbsbshxhxhdhd
3 0
3 years ago
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