reallize. ALCEBRA//FOUNOATIONAL SKLUS PERIOD $$ 4 / 8 $$ (IMS) 5-3: MathXL for Schoot: Additional Practice (LMS graded) An unusually wet spring has caused the size of a mosquito population in some city to increase by $$ 6 \% $$ each day. If an estimated 190.000 mosquitoes are in the city on May 6, find how many mosquitoes will inhabit the city on May 19. Use $$ y=190,000(1.06)^{x} $$ where $$ x $$ is number of days since May 6. $$ \qquad $$ On May 19 there will be approximately $$ \square $$ mosquitoes. (Round to the nearest thousand as needed.)

Question

UpStudy AI · Accepted Answer

To determine the number of mosquitoes inhabiting the city on May 19, we'll use the provided exponential growth model:

$$ y = 190,000 \times (1.06)^x $$

where:
- $$ y $$ = number of mosquitoes
- $$ x $$ = number of days since May 6

**Step 1: Calculate the Number of Days**

First, determine the number of days between May 6 and May 19.

- **May 6** to **May 19** is **13 days**.

So, $$ x = 13 $$.

**Step 2: Plug the Values into the Formula**

$$ y = 190,000 \times (1.06)^{13} $$

**Step 3: Calculate $$ (1.06)^{13} $$**

Let's compute $$ (1.06)^{13} $$:

$$
\begin{align*}
(1.06)^1 &= 1.06 \
(1.06)^2 &= 1.06 \times 1.06 = 1.1236 \
(1.06)^3 &= 1.1236 \times 1.06 \approx 1.191016 \
\vdots \
(1.06)^{13} &\approx 2.129466
\end{align*}
$$

**Step 4: Multiply to Find $$ y $$**

$$
y = 190,000 \times 2.129466 \approx 404,598.54
$$

**Step 5: Round to the Nearest Thousand**

$$
404,598.54 \approx 405,000 \text{ mosquitoes}
$$

**Conclusion**

On May 19, there will be approximately **405,000 mosquitoes** inhabiting the city.
On May 19, there will be approximately 405,000 mosquitoes in the city.

Solución de tutoría real

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Algebra Questions