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4 years ago

Find the median, first quartile, third quartile, and the interquartile of the data set.

Mathematics

1 answer:

Schach [20]4 years ago

3 0

Answer:

54 = median

48 = 1st quartile

69 = 3rd quartile

21 = interquartile

Hope this helped!! :)

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3 years ago

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3 years ago

Read 2 more answers

In a dog show, there will be a gold and silver medal awarded.There are 6 finalists. In how many ways can the 2 medals be awarded

Mandarinka [93]

Answer:

Step-by-step explanation:

Given: There are six finalists and two medals. That is it asks for the possibilities of 6 finalists taking two at a time. Say, for example the first and third finalist or the second and the sixth finalist and so on. The number of possibilities is given by $^nP_r$

$^nP_r = \frac{n!}{(n - r)!}$

So the number of possibilities of 6 finalists taken two at a time is $^6P_2$

$\implies \frac{6!}{(6 - 2)!} = \frac{6!}{4!} \hspace{15mm} [n! = n \times (n-1) \times (n - 2) \times . . . 2 \times 1]$

$\implies \frac{6 \times 5 \times 4 \times \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 6 \times 5 = 30$

4 0

4 years ago

Helppp nowwww plssss!!!

Radda [10]

Answer
the least common is 21
3x7=21 7x3=21
1/3=7/21
5/7= 15/21

hope this helps and have a wonderful day :)

7 0

3 years ago

Please help me with this question!

frez [133]

$\text {Laura = } 12 \text { years } 3 \text { months }$

$\text {Lisa = } 11 \text { years } 7 \text { months }$

$\text {Deborah = } 11 \dfrac{1}{2} \text { years } = 11 \text { years } 6 \text { months }$

$\text {Elizabeth = } 11 \dfrac{1}{3} \text { years } = 11 \text { years } 4 \text { months }$

-----------------------------------------------------
Find Total :
-----------------------------------------------------
$\text {Total = } 12 \text { yr } 3 \text { mth } + 11 \text { yr } 7 \text { mth } + 11 \text { yr} 6 \text { mth} +11 \text { yr } 4 \text { mth}$

$\text {Total = } 45 \text { years } 20 \text { months }$

-----------------------------------------------------
Find Average :
-----------------------------------------------------
$\text {Average = } (45 \text { years } 20 \text { months }) \div 4$

$\text {Average = } 560 \text { months } \div 4$

$\text {Average = } 140 \text { months } \div 4$

$\text {Average = } 11 \text { years } 8 \text { months }$

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$\bf \text {Answer : Average = } 11 \text { years } 8 \text { months }$
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6 0

4 years ago