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Elina [12.6K]
3 years ago
9

Carol picked 8 quarts of strawberry and paid $10.00 for them. How much did she pay per quart? Write a proportion

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
6 0

Answer:

the answer is 0.4

Step-by-step explanation:

notka56 [123]3 years ago
6 0
Money paid for 8 quarts of strawberry = $10.00.
Money paid for each quart = ?
8 divided by 10.00=1.25
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Ruben bought 2 large popcorn buckets and 4 small drinks for $20. Lucia bought 2 large popcorn buckets and 2 small drinks for $14
soldier1979 [14.2K]

Answer:

$3 for the small drinks and $4 for the large popcorn

Step-by-step explanation:

so we know that 2 popcorns and 4 small drinks give us $20 and 2 popcorns and 2 drinks gives us 14. so $6 was taken away because of two drinks and 6÷2 equals 3 therefore $3 is how much a small drink costs. after taking away another 6 from 14 because we found the costs of drinks we get $8 for two popcorns and 8÷2 is 4 so a large popcorn costs $4

4 0
2 years ago
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Find the value of x in rectangle BCDE.<br> Find Value of x.
Schach [20]

Answer:

34

Step-by-step explanation:

6 0
2 years ago
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In the New Orleans metropolitan area in August​ 2005, the labor force was​ 634,512 and​ 35,222 people were unemployed. In Septem
Lisa [10]

Answer:

 Unemployment (August) = 35'222/634'512 ≈ 0.0555 ≈ 5.55%

Employed (August) = 634'512-35'222 = 599'290

Labor force (September) = 634'512-156'518 = 477'994

Employed (September) = 599'290-206'024 = 393'266

Unemployed (September) = 477'994-393'266 = 84'728

Unemployment (September) = 84'728/477'994 ≈ 0.177 ≈ 17.7%

Step-by-step explanation:

8 0
3 years ago
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Virty [35]

Answer:

20%-30%

Step-by-step explanation:

KJXVKGHJFSVTFKTJKHFDSYU rifixbxiifxojfjfvshssjhssjwkhxxc

4 0
2 years ago
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The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000​
Papessa [141]

Answer:

here i finished!

hope it helps yw!

Step-by-step explanation:

The doubling period of a bacterial population is 15 minutes.

At time t = 90 minutes, the bacterial population was 50000.

Round your answers to at least 1 decimal place.

:

We can use the formula:

A = Ao*2^(t/d); where:

A = amt after t time

Ao = initial amt (t=0)

t = time period in question

d = doubling time of substance

In our problem

d = 15 min

t = 90 min

A = 50000

What was the initial population at time t = 0

Ao * 2^(90/15) = 50000

Ao * 2^6 = 50000

We know 2^6 = 64

64(Ao) = 50000

Ao = 50000/64

Ao = 781.25 is the initial population

:

Find the size of the bacterial population after 4 hours

Change 4 hr to 240 min

A = 781.25 * 2^(240/15

A = 781.25 * 2^16

A= 781.25 * 65536

A = 51,199,218.75 after 4 hrs

6 0
3 years ago
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