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ASHA 777 [7]
3 years ago
10

What is 2.30+30= 5.30 True or false or2.60 true or false ​

Mathematics
2 answers:
Kruka [31]3 years ago
5 0

Answer:

false

Step-by-step explanation:

2.30+30= 32.30

son4ous [18]3 years ago
4 0
<h2>Answer:  Step-by-step explanation:   2.30+30= 5.30 -FALSE!!  2.30+30=2.60 -FALSE!!    REAL ANSWER: 32.3</h2>
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There are 24 pieces of fruit in a bowl and 20 of them are apples. What percentage of the pieces of fruit in the bowl are apples?
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Answer:

83.3%

Step-by-step explanation:

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3 years ago
Estimate the sum of 196 and 482
fredd [130]

we know that

A quick way to estimate the sum of two numbers is to round each number and then add the rounded numbers

case a) We round the number to the nearest hundred

so

196 Round up is equal to 200

482 Round up is equal to 500

Find the estimate sum

200+500=700

case b) We round the number to the nearest tens

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Number 6. is 34. because 67 - 34 is 33
3 0
3 years ago
What is another name for vertices?
BlackZzzverrR [31]

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Apogee

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5 0
3 years ago
Hospital sells raffle tickets to raise fund for new medical equipment. Last year, 2000 tickets were sold for $24 each. The fund-
Zanzabum

Answer:

1) Price decrease = $4; 2) new price = $20; 3) maximum revenue = $50 000

Step-by-step explanation:

The hospital sold 2000 tickets for $24 each

Revenue = price per ticket × number of tickets sold

Let x = change in price

New price = 24 - x

New number of tickets sold = 2000 + 125x

1) Calculate change in price to maximize revenue

y = (24 - x)(2000 + 125x)

y = 48 000 + 1000x - 125x²

y = -125x² + 1000x + 48 000

a = -125; b = 1000; c = 48 000

The vertex is at

x = -\dfrac{b}{2a} = -\dfrac{1000}{2(-125)}= \dfrac{1000}{250} = \mathbf{4}

A price decrease of $4 will maximize revenue.

2) New ticket price

Original price = $24

Price change =  <u>  - 4 </u>

New price =       $20

A ticket price of $20 will maximize revenue.

3) Maximum revenue

         Ticket price = $20

No of tickets sold = 2000 + 125(4) = 2000 + 500  = 2500

             Revenue = 2500 × $20 = $ 50 000

The maximum revenue is $50 000.

The graph below slows the relation between the price drop and total revenue.  A price drop of $4 results in a maximum revenue of $50 000.

8 0
3 years ago
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