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Dmitriy789 [7]
4 years ago
12

It take you 0.8 of a minute to read each page of your health book. It takes you 5.5 minutes to take the test at the end. How lon

g will it take you to read 6.25 pages and also take the test? Is it 10.5 or 11.75 or 12.55 or 39.375 minutes?
Mathematics
2 answers:
guapka [62]4 years ago
6 0
So,we know that it take 0.8 of a minute or 48 seconds to read each page of a health book, so it will take 300 seconds to read 6.25 pages. As a result, it will take 10.5 minutes to finish reading 6.25 pages and also taking a test. Hope it help!
jarptica [38.1K]4 years ago
3 0

Answer:

39.375 minutes

Step-by-step explanation:

Since, each test on a page takes 5.5 minutes while each page take 0.8 minutes to read. So, for 6.25 pages. So, to give the test of 6.25 pages, time will be:

(6.25).(5.5)

34.375 minutes.

Now, to read these 6.25 pages, time will be:

(6.25).(0.8)

5 minutes

So, the total time taken is

34.375 + 5

39.375 minutes

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Find the common denominator 2/7 and 7/10
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2 years ago
Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of
ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

A(t) = A(0)(1-r)^t

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

Hourly rate of change:

Decreases 26% by day. A day has 24 hours. This means that A(24) = (1-0.26)A(0) = 0.74A(0); We use this to find r.

A(t) = A(0)(1-r)^t

0.74A(0) = A(0)(1-r)^{24}

(1-r)^{24} = 0.74

\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}

1 - r = (0.74)^{\frac{1}{24}}

1 - r = 0.9875

r = 1 - 0.9875 = 0.0125

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that A(0) = 810

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

A(t) = A(0)(1-r)^t

A(t) = 810(1 - 0.26)^t

A(t) = 810(0.74)^t

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

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3 years ago
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