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natka813 [3]
3 years ago
9

What is the median of the set of data

Mathematics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

The median average is the middle number in a set of data , when the data has been written in ascending size order. If there is an even number of items of data, there will be two numbers in the middle. The median is the number that is half way between these two numbers.

I hope it's helpful!

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ms.Alonzo ordered 4,000 markers for her store. only one tenth of them arrived . how many markers did she receive
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Ms. Alonzo received 400 markers because one tenth of 4000 is 400. Hope this helped! 
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3 years ago
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The container that holds the water for the football team is 2/3
Kipish [7]

Answer:

18 gallons

Step-by-step explanation:

We'll find out how much of the total container was lost when pouring out by subtracting 2/3 from 1/2.

2/3 = 4/6

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4/6 - 3/6 = 1/6

1/6 of the total amount the container could hold was lost when 3 gallons was poured out. This means that 1/6 of the container equals to 3 gallons. We can then simply multiply 3 by 6 to find the number of gallons of liquid the container can hold.

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Checking my answer:

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This means that our answer is correct.

8 0
3 years ago
Help me plz i’m posting rn
nordsb [41]

Answer:

y = 7

Step-by-step explanation:

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3 0
2 years ago
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A case of tomato cans weighs 563 dekagrams. A case of soup cans weighs 458 dekagrams. How much do the two cases weigh together i
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Answer:

  102,100 decigrams

Step-by-step explanation:

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The two cases together weigh 102,100 decigrams.

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6 0
3 years ago
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4 Tan A/1-Tan^4=Tan2A + Sin2A​
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tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
2 years ago
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