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Allushta [10]
3 years ago
15

Plz help!! asap!!

Mathematics
2 answers:
Dima020 [189]3 years ago
6 0
Sale price is $44.80
Discount is $19.20
64/100=0.64
0.64*30=19.20
64-19.20=44.80
Marat540 [252]3 years ago
5 0
Multiply 64 times .30 =?
Then take the answer you get and subtract it from $64 then you should get your answer of $44.80 please work it out it’s not too hard it is actually pretty easy.
Hope this helps
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44 kg of onions cost $308. How much would 30 kg cost?
maks197457 [2]

Answer:

$210

Step-by-step explanation:

Cost of 44kg onions = $308

Cost of 1kg onion = \frac{308}{44}  = 7

Cost of 30kg onions = 30 \times 7 = 210

7 0
2 years ago
Ben consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Ben's body decr
Airida [17]

Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

A(t) = A(0)e^{-kt}

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, A(10) = 0.7278A(0). We use this to find k.

A(t) = A(0)e^{-kt}

0.7278A(0) = A(0)e^{-10k}

e^{-10k} = 0.7278

\ln{e^{-10k}} = \ln{0.7278}

-10k = \ln{0.7278}

k = -\frac{\ln{0.7278}}{10}

k = 0.03177289938


Then

A(t) = A(0)e^{-0.03177289938t}

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

A(5) = A(0)e^{-0.03177289938*5}

A(5) = 0.8531

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

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3 years ago
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Answer:

x=15, y=19

Step-by-step explanation:

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3 years ago
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3 years ago
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Please help........................
Art [367]

Answer:

I think it is 2 one

Step-by-step explanation:

7 0
3 years ago
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