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3 years ago

Original price:$29.99;markdown: 33 1/3% Find the retail price

Mathematics

2 answers:

faust18 [17]3 years ago

5 0

Answer:

Retail Price of the item is $ 19.99 (approx.)

Step-by-step explanation:

Given:

Original Price of an item = $ 29.99

Markdown Percentage = $33\frac{1}{3}$ % = $\frac{33\times3+1}{3}=\frac{100}{3}$ %

We need to find: Retail Price

We know that Retail price = Original Price - Markdown amount

Markdown amount = $\frac{\frac{100}{3}}{100}\times29.99$

= $\frac{100}{300}\times29.99$

= $\frac{2999}{300}$

= $ 10 (approx.)

So, Retail price = 29.99 - 10 = $ 19.99

Therefore, Retail Price of the item is $ 19.99 (approx.)

-Dominant- [34]3 years ago

3 0

33.3%, move the decimal to the left twice to make it easier. then multiply 29.99 by .333. take that and subtract it from 29.99. I got 20, I'm not sure if that's right though

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he radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the bones from a mastodon had lost

elena-s [515]

Answer:

The bones were 12,485 years old at the time they were discovered.

Step-by-step explanation:

Amount of the element:

The amount of the element after t years is given by the following equation, considering the decay rate proportional to the amount present:

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

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

The radioactive element carbon-14 has a half-life of 5750 years.

This means that $A(5750) = 0.5A(0)$ , and we use this to find k. So

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

$0.5A(0) = A(0)e^{-5750k}$

$e^{-5750k} = 0.5$

$\ln{e^{-5750k}} = \ln{0.5}$

$-5750k = \ln{0.5}$

$k = -\frac{\ln{0.5}}{5750}$

$k = 0.00012054733$

$A(t) = A(0)e^{-0.00012054733t}$

A scientist determined that the bones from a mastodon had lost 77.8 % of their carbon-14. How old were the bones at the time they were discovered?

Had 100 - 77.8 = 22.2% remaining, so this is t for which:

$A(t) = 0.222A(0)$