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kondaur [170]
3 years ago
15

What is the Median for the following set of numbers? ​21 23 76 47 55 135 45 30 17

Mathematics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

Median = 45

Step-by-step explanation:

We are given the following data set:

21, 23, 76, 47, 55, 135, 45, 30, 17

Median is the number that divides the data into two equal parts.

Formula:

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Sorted data:

17, 21, 23, 30, 45, 47, 55, 76, 135

Sample size = 9, which is odd

Median =

\dfrac{9+1}{2}^{th}\text{ term} = \dfrac{10}{2}^{th}\text{ term} = 5^{th}\text{ term}\\\\= 45

The median of given set of numbers is 45.

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Godfrey plays a game in which he throws two fair six sided dice. If he rolls two sizes, he wins 20p, if he rolls one six, he win
blagie [28]

Answer:

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Step-by-step explanation:

If random variable X represents the amount Godfrey gains in one turn.

There are 3 different possible outcomes for X.

- Godfrey pays 5p to enter the game and gets two sixes and wins 20p.

Net gain = 15p

Probability of getting two sixes from two fair dice

= (number of outcomes with two sixes) ÷ (total number of outcomes)

number of outcomes with two sixes = 1

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (1/36) = 0.0278

- Godfrey pays 5p to enter the game and gets only one six and wins 10p.

Net gain = 5p

Probability of getting one six from either of two fair dice

= (number of outcomes with one six) ÷ (total number of outcomes)

number of outcomes with one six = 2 × n[(6,1), (6,2), (6,3), (6,4), (6,5)] = 2 × 5 = 10

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (10/36) = 0.278

- Godfrey pays 5p to enter the game and doesn't win anything

Net gain = -5p

Probability of not getting two sixes or one six.

= 1 - [(Probability of getting two sixes) + (Probability of getting one six on.wither dice)]

= 1 - 0.0278 - 0.278 = 0.6942

Probability of getting not getting two sixes or one six = 0.6942

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Hope this Helps!!!

4 0
3 years ago
Read 2 more answers
A case of Mountain Dew (24 cans) cost $7.68. What is the unit price?
Westkost [7]
0.32 cents for each unit because if you divide 7.68 by 24 it is 0.32
3 0
3 years ago
Read 2 more answers
I need answer. Please help me.​
Vanyuwa [196]

I think it’s positive. As it gets heavier, the cost has a positive slope, or increases. I’m not absolutely sure though

6 0
3 years ago
Read 2 more answers
There are four movies showing at the theater, 2/5 of the people bought tickets for the comedy, 1/4 bought tickets for the horror
RUDIKE [14]

Given:

Number of people bought tickets for comedy = \frac{2}{5}

Number of people bought tickets for horror = \frac{1}{4}

Number of people bought tickets for kids movie = \frac{3}{10}

To find the number of people who bought tickets for the action movie.

Let us take,

Total number of tickets = 1

Now,

Total number of tickets for comedy, horror and kids movie are = (\frac{2}{5} +\frac{1}{4} +\frac{3}{10} )

So,

(\frac{2}{5} +\frac{1}{4} +\frac{3}{10} )

LCM of 5,4,10 = 20

= \frac{8+5+6}{20}

= \frac{19}{20}

Rest are action movie's tickets.

Therefore,

Number of action movie tickets = 1-\frac{19}{20}

= \frac{20-19}{20}

= \frac{1}{20}

Hence,

The number of people who bought the tickets for action movie is \frac{1}{20}.

7 0
3 years ago
Suppose that 3 is a factor of a, a is a divisor of 12, and a is positive. What is the number of possible values of a?
Zina [86]
Answer: the number of possible values of a are 3 and 6

Explanation:
Since a is a divisor of 12
Then let’s see what are the divisor for 12
12= 6 x 2
12= 3 x 4
And we know that 3 is a factor of a
So if a was 3 then 3/3 is exactly 1
And if a was 6 then 6/3 is exactly 2 with no remainder

And 3 and 6 are also the divisor of 12
12/6=2
12/3=4
So a should be 3 and 6

5 0
3 years ago
Read 2 more answers
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