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2 years ago

The half life of a chemical element substance is 16 days, How long will it takes for the substance to decay from 120mg to 2 mg

Mathematics

1 answer:

Studentka2010 [4]2 years ago

8 0

Answer:

94.5 days

Step-by-step explanation:

We are asked to calculate the time.

The formula to calculate the time for half life =

t = t½ × In (Nt/No)/-In2

t½ = half life = 16 days

No = Initial substance = 120mg

Nt = Amount of substances after time t = 2 mg

t = 16 × In (2/120)/-In2

t = 94.510249529736 days

Approximately = 94.5 days

Therefore, the time it would take for the substance to decay from 120mg to 2 mg is 94.5 days.

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2 years ago

Read 2 more answers

A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$

Upon substituting C's value from equation (2) in equation (1) we will get,

$\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)$

Let us have a common denominator on right side of equation.

$\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)$

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Multiplying both sides of our equation by 7, we will get,

$7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)$

$C+D-10=\frac{7}{11}*(C+D-2)$

Multiplying both sides of our equation by 11, we will get,

$11*(C+D-10)=11*\frac{7}{11}*(C+D-2)$

$11*(C+D-10)=7*(C+D-2)$

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$11C-7C+11D-7D=-14+110$

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2 years ago

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Answer:

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The value of the car is decreasing. Thus, by the end of the first year of ownership, the car will have the value $28,000(1 -0.189)^1, where the "-0.189" represents DECAY instead of GROWTH.

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2 years ago