Math / Algebra

QuestionThe following exponential function models the number of students who are ill with Covid as a result of infection by other students in a large lecture section: $N(t)=3\left(1.3^{\frac{t}{4}}\right)$ $N$ is the number of infected students and $t$ is the number of days after the initial exposure. Answer the following questions:
Is this a growth or decay function? Type growth or decay: $\square$ A
The percentage increase in Covid cases is $\square$ A in how many days? $\square$ A
How many students were infected initially? $\square$ $\Delta$

Studdy Solution

STEP 1

What is this asking? We're given a formula that tells us how many students are sick with Covid after some number of days, and we need to figure out if the sickness is spreading or shrinking, how fast it's changing, and how many students were sick at the very beginning. Watch out! The formula isn't in the simplest form, so we need to be careful when figuring out the percentage increase and over how many days.
Don't mix up the initial number of sick students with the rate of change!

STEP 2

1. Determine if the function represents growth or decay.
2. Find the percentage increase and the period over which it occurs.
3. Calculate the initial number of infected students.

STEP 3

Alright, let's look at the formula: $N(t)=3\left(1.3^{\frac{t}{4}}\right)$ .
The key part here is the base of the exponent, which is 1.3.

STEP 4

Since 1.3 is greater than 1, this tells us we have exponential growth!
It's like a chain reaction, where each sick student infects more than one new student, causing the number of sick students to increase rapidly.

STEP 5

Let's rewrite the base, 1.3, as $1 + 0.3$ .
Remember, adding to one is how we represent growth!
So, $N(t)=3\left((1 + 0.3)^{\frac{t}{4}}\right)$ .

STEP 6

This 0.3 represents the rate of growth as a decimal.
To get the percentage increase, we multiply it by 100: 0.3 \cdot 100 = \textbf{30%}.
So, the number of sick students is increasing by 30%.

STEP 7

Now, notice that the exponent is $\frac{t}{4}$ .
This means it takes 4 days for the 30% increase to happen.
If the exponent were just $t$ , then the increase would happen every day.
But since it's $\frac{t}{4}$ , the 30% increase happens every 4 days.

STEP 8

To find the initial number of infected students, we need to find the value of $N(t)$ when $t = 0$ , which represents the very beginning.
So, we substitute $t = 0$ into our formula: $N(0) = 3\left(1.3^{\frac{0}{4}}\right)$ .

STEP 9

This simplifies to $N(0) = 3\left(1.3^0\right)$ .
Remember, any number (except 0) raised to the power of 0 is equal to 1!

STEP 10

So, $N(0) = 3 \cdot 1 = \textbf{3}$ .
This means there were 3 students infected initially.

STEP 11

The function represents growth.
There is a 30% increase in Covid cases every 4 days.
Initially, 3 students were infected.

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Studdy Solution

STEP 1

STEP 2

1. Determine if the function represents growth or decay.
2. Find the percentage increase and the period over which it occurs.
3. Calculate the initial number of infected students.

STEP 3

Alright, let's look at the formula: $N(t)=3\left(1.3^{\frac{t}{4}}\right)$ .
The key part here is the base of the exponent, which is 1.3.

STEP 4

Since 1.3 is greater than 1, this tells us we have exponential growth!
It's like a chain reaction, where each sick student infects more than one new student, causing the number of sick students to increase rapidly.

STEP 5

Let's rewrite the base, 1.3, as $1 + 0.3$ .
Remember, adding to one is how we represent growth!
So, $N(t)=3\left((1 + 0.3)^{\frac{t}{4}}\right)$ .

STEP 6

This 0.3 represents the rate of growth as a decimal.
To get the percentage increase, we multiply it by 100: 0.3 \cdot 100 = \textbf{30%}.
So, the number of sick students is increasing by 30%.

STEP 7

Now, notice that the exponent is $\frac{t}{4}$ .
This means it takes 4 days for the 30% increase to happen.
If the exponent were just $t$ , then the increase would happen every day.
But since it's $\frac{t}{4}$ , the 30% increase happens every 4 days.

STEP 8

To find the initial number of infected students, we need to find the value of $N(t)$ when $t = 0$ , which represents the very beginning.
So, we substitute $t = 0$ into our formula: $N(0) = 3\left(1.3^{\frac{0}{4}}\right)$ .

STEP 9

This simplifies to $N(0) = 3\left(1.3^0\right)$ .
Remember, any number (except 0) raised to the power of 0 is equal to 1!

STEP 10

So, $N(0) = 3 \cdot 1 = \textbf{3}$ .
This means there were 3 students infected initially.

STEP 11

The function represents growth.
There is a 30% increase in Covid cases every 4 days.
Initially, 3 students were infected.

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Studdy Solution

STEP 1

STEP 2

1. Determine if the function represents growth or decay.2. Find the percentage increase and the period over which it occurs.3. Calculate the initial number of infected students.

STEP 3

Alright, let's look at the formula: N(t)=3(1.3t4)N(t)=3\left(1.3^{\frac{t}{4}}\right)N(t)=3(1.34t​).The **key part** here is the base of the exponent, which is **1.3**.

STEP 4

Since **1.3** is *greater* than 1, this tells us we have *exponential growth*!It's like a chain reaction, where each sick student infects more than one new student, causing the number of sick students to increase rapidly.

STEP 5

Let's **rewrite** the base, 1.3, as 1+0.31 + 0.31+0.3.Remember, adding to one is how we represent growth!So, N(t)=3((1+0.3)t4)N(t)=3\left((1 + 0.3)^{\frac{t}{4}}\right)N(t)=3((1+0.3)4t​).

STEP 6

This **0.3** represents the rate of growth as a decimal.To get the **percentage increase**, we multiply it by 100: 0.3 \cdot 100 = \textbf{30%}.So, the number of sick students is increasing by **30%**.

STEP 7

Now, notice that the exponent is t4\frac{t}{4}4t​.This means it takes **4** days for the **30%** increase to happen.If the exponent were just ttt, then the increase would happen every day.But since it's t4\frac{t}{4}4t​, the **30%** increase happens every **4** days.

STEP 8

To find the initial number of infected students, we need to find the value of N(t)N(t)N(t) when t=0t = 0t=0, which represents the very beginning.So, we **substitute** t=0t = 0t=0 into our formula: N(0)=3(1.304)N(0) = 3\left(1.3^{\frac{0}{4}}\right)N(0)=3(1.340​).

STEP 9

This simplifies to N(0)=3(1.30)N(0) = 3\left(1.3^0\right)N(0)=3(1.30).Remember, *any* number (except 0) raised to the power of 0 is equal to 1!

STEP 10

So, N(0)=3⋅1=3N(0) = 3 \cdot 1 = \textbf{3}N(0)=3⋅1=3.This means there were **3** students infected initially.

STEP 11

The function represents *growth*.There is a **30%** increase in Covid cases every **4** days.Initially, **3** students were infected.

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1. Determine if the function represents growth or decay.
2. Find the percentage increase and the period over which it occurs.
3. Calculate the initial number of infected students.

Alright, let's look at the formula: $N(t)=3\left(1.3^{\frac{t}{4}}\right)$ .
The key part here is the base of the exponent, which is 1.3.

Since 1.3 is greater than 1, this tells us we have exponential growth!
It's like a chain reaction, where each sick student infects more than one new student, causing the number of sick students to increase rapidly.

Let's rewrite the base, 1.3, as $1 + 0.3$ .
Remember, adding to one is how we represent growth!
So, $N(t)=3\left((1 + 0.3)^{\frac{t}{4}}\right)$ .

This 0.3 represents the rate of growth as a decimal.
To get the percentage increase, we multiply it by 100: 0.3 \cdot 100 = \textbf{30%}.
So, the number of sick students is increasing by 30%.

Now, notice that the exponent is $\frac{t}{4}$ .
This means it takes 4 days for the 30% increase to happen.
If the exponent were just $t$ , then the increase would happen every day.
But since it's $\frac{t}{4}$ , the 30% increase happens every 4 days.

To find the initial number of infected students, we need to find the value of $N(t)$ when $t = 0$ , which represents the very beginning.
So, we substitute $t = 0$ into our formula: $N(0) = 3\left(1.3^{\frac{0}{4}}\right)$ .

This simplifies to $N(0) = 3\left(1.3^0\right)$ .
Remember, any number (except 0) raised to the power of 0 is equal to 1!

So, $N(0) = 3 \cdot 1 = \textbf{3}$ .
This means there were 3 students infected initially.

The function represents growth.
There is a 30% increase in Covid cases every 4 days.
Initially, 3 students were infected.