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Common Core: High School - Functions : Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

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Example Question #1 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An anti-diabetic drug, has a half-life of about hours. If a patient was administered $200 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=184\textup{ mg}$

$A=82\textup{ mg}$

$A=90\textup{ mg}$

$A=94\textup{ mg}$

$A=84\textup{ mg}$

Correct answer:

$A=84\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

For the purpose of Common Core Standards, recognize situations that have a exponential growth or decay over a certain interval, falls within the Cluster A of construct and compare linear, quadratic, and exponential model and solve problems concept (CCSS.Math.content.HSF.LE.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=200\textup{ mg} \\h=\textup{half life}=4\textup{ hours} \\t_f=1:00\textup{ pm} \\t_i=8:00\textup{ am} \\A=?$

Step 2: Calculate .

$1:00-8:00\rightarrow 5 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=200\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{4}}$

$A=200\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.25}$

$A=200\textup{ mg}\cdot 0.420$

$A=84\textup{ mg}$

Report an Error

Example Question #2 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An anti-diabetic drug, has a half-life of about 6.2 hours. If a patient was administered $500 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=159.57\textup{ mg}$

$A=199.99\textup{ mg}$

$A=197.57\textup{ mg}$

$A=128.57\textup{ mg}$

$A=199.57\textup{ mg}$

Correct answer:

$A=199.57\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=500\textup{ mg} \\h=\textup{half life}=6.2\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=500\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{6.2}}$

$A=500\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.325}$

$A=500\textup{ mg}\cdot 0.399$

$A=199.57\textup{ mg}$

Report an Error

Example Question #3 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An particular medicine, has a half-life of about hours. If a patient was administered $50 mg of the drug at $11 am , how much is left at $2 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=45\textup{ mg}$

$A=20\textup{ mg}$

$A=15\textup{ mg}$

$A=30\textup{ mg}$

$A=25\textup{ mg}$

Correct answer:

$A=25\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=50\textup{ mg} \\h=\textup{half life}=3\textup{ hours} \\t_f=2:00\textup{ pm} \\t_i=11:00\textup{ am} \\A=?$

Step 2: Calculate .

$2:00-11:00\rightarrow 3 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=50\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{3}{3}}$

$A=50\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1}$

$A=50\textup{ mg}\cdot 0.50$

$A=25\textup{ mg}$

Report an Error

Example Question #4 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An particular medicine, has a half-life of about hours. If a patient was administered $600 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=266\textup{ mg}$

$A=366\textup{ mg}$

$A=360\textup{ mg}$

$A=306\textup{ mg}$

$A=466\textup{ mg}$

Correct answer:

$A=366\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=600\textup{ mg} \\h=\textup{half life}=7\textup{ hours} \\t_f=1:00\textup{ pm} \\t_i=8:00\textup{ am} \\A=?$

Step 2: Calculate .

$1:00-8:00\rightarrow 5 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{7}}$

$A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{0.714}$

$A=600\textup{ mg}\cdot 0.61$

$A=366\textup{ mg}$

Report an Error

Example Question #5 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An particular medicine, has a half-life of about hours. If a patient was administered $350 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=197.35\textup{ mg}$

$A=196.35\textup{ mg}$

$A=169.35\textup{ mg}$

$A=195.65\textup{ mg}$

$A=198.75\textup{ mg}$

Correct answer:

$A=196.35\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=350\textup{ mg} \\h=\textup{half life}=6\textup{ hours} \\t_f=1:00\textup{ pm} \\t_i=8:00\textup{ am} \\A=?$

Step 2: Calculate .

$1:00-8:00\rightarrow 5 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=350\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{6}}$

$A=350\textup{ mg}\cdot \left(\frac{1}{2}\right)^{0.833}$

$A=350\textup{ mg}\cdot 0.561$

$A=196.35\textup{ mg}$

Report an Error

Example Question #6 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An anti-diabetic drug, has a half-life of about 6.2 hours. If a patient was administered $800 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=318.2\textup{ mg}$

$A=321.2\textup{ mg}$

$A=319.2\textup{ mg}$

$A=320.2\textup{ mg}$

$A=317.3\textup{ mg}$

Correct answer:

$A=319.2\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=800\textup{ mg} \\h=\textup{half life}=6.2\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=800\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{6.2}}$

$A=800\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.325}$

$A=800\textup{ mg}\cdot 0.399$

$A=319.2\textup{ mg}$

Report an Error

Example Question #7 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An anti-diabetic drug, has a half-life of about 6.2 hours. If a patient was administered $50 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=17.95\textup{ mg}$

$A=19.97\textup{ mg}$

$A=18.95\textup{ mg}$

$A=21.95\textup{ mg}$

$A=19.95\textup{ mg}$

Correct answer:

$A=19.95\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=50\textup{ mg} \\h=\textup{half life}=6.2\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=50\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{6.2}}$

$A=50\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.325}$

$A=50\textup{ mg}\cdot 0.399$

$A=19.95\textup{ mg}$

Report an Error

Example Question #8 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An medicine, has a half-life of about 4.2 hours. If a patient was administered $100 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=27.77\textup{ mg}$

$A=35.77\textup{ mg}$

$A=24.57\textup{ mg}$

$A=25.57\textup{ mg}$

$A=25.77\textup{ mg}$

Correct answer:

$A=25.77\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=100\textup{ mg} \\h=\textup{half life}=4.2\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=100\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{4.2}}$

$A=100\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.9563}$

$A=100\textup{ mg}\cdot 0.2577$

$A=25.77\textup{ mg}$

Report an Error

Example Question #9 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An certain medicine, has a half-life of about 3.4 hours. If a patient was administered $150 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=29.08\textup{ mg}$

$A=29.09\textup{ mg}$

$A=28.09\textup{ mg}$

$A=28.08\textup{ mg}$

$A=28.39\textup{ mg}$

Correct answer:

$A=28.09\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=150\textup{ mg} \\h=\textup{half life}=3.4\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=150\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{3.4}}$

$A=150\textup{ mg}\cdot \left(\frac{1}{2}\right)^{2.4166}$

$A=150\textup{ mg}\cdot 0.1873$

$A=28.09\textup{ mg}$

Report an Error

Example Question #10 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

An particular medicine, has a half-life of about hours. If a patient was administered $800 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Possible Answers:

$A=517.82\textup{ mg}$

$A=520.72\textup{ mg}$

$A=516.72\textup{ mg}$

$A=518.02\textup{ mg}$

$A=518.72\textup{ mg}$

Correct answer:

$A=518.72\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=800\textup{ mg} \\h=\textup{half life}=8\textup{ hours} \\t_f=1:00\textup{ pm} \\t_i=8:00\textup{ am} \\A=?$

Step 2: Calculate .

$1:00-8:00\rightarrow 5 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=800\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{8}}$

$A=800\textup{ mg}\cdot \left(\frac{1}{2}\right)^{0.625}$

$A=800\textup{ mg}\cdot 0.6484$

$A=518.72\textup{ mg}$

Report an Error

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Common Core: High School - Functions : Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #1 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #2 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #3 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #4 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #5 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #6 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #7 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #8 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #9 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

Example Question #10 : Growth And Decay By Contant Percent Rate: Ccss.Math.Content.Hsf Le.A.1c

All Common Core: High School - Functions Resources

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