Question

11 a company opens an account with an initial balance of $36,100.00. the account earns interest, and no additional deposits or withdrawals are made. the account balance is given by an exponential function a, where a(t) is the account balance, in dollars, t years after the account is opened. the account balance after 13 years is $68,071.93. which equation could define a? a) a(t)=36,100.00(1.05)^t b) a(t)=31,971.93(1.05)^t c) a(t)=31,971.93(0.05)^t d) a(t)=36,100.00(0.05)^t 12 the circle shown has center o, circumference 144π. note figure not drawn to scale. 13 y = -2x 3x + y = 40 the solution to the given system of equations is (x,y). what is the value of x? 14 data value frequency 6 3 7 3 8 8 9 8 10 9 11 11 12 9 13 0 14 6 the frequency table summarizes the 57 data values in a data set. what is the maximum data value in the data set?

Explanation:

11.

Step1: Recall compound - interest formula

The general form of an exponential - growth function for compound interest is $A(t)=P(1 + r)^t$, where $P$ is the initial amount, $r$ is the annual interest rate, and $t$ is the number of years. Here, $P = 36100$.

Step2: Test each option

We know that when $t = 13$, $A(13)$ should be approximately $68071.93$.
For option A: $A(t)=36100(1.05)^t$. When $t = 13$, $A(13)=36100\times(1.05)^{13}$.
Using the formula $a^n=e^{n\ln(a)}$, $(1.05)^{13}=e^{13\ln(1.05)}\approx1.885649$. Then $A(13)=36100\times1.885649\approx68071.93$.
For option B: The initial amount $P = 31971.93
eq36100$, so it's incorrect.
For option C: The base of the exponential function $(0.05)^t$ represents decay, not growth, so it's incorrect.
For option D: The base of the exponential function $(0.05)^t$ represents decay, not growth, so it's incorrect.

Step1: Substitute $y=-2x$ into the second equation

Given $y = - 2x$ and $3x + y=40$, substitute $y$ in the second - equation: $3x+( - 2x)=40$.

Step2: Simplify the left - hand side of the equation

$3x-2x=x$. So, $x = 40$.

The frequency table lists data values and their corresponding frequencies. The data values are listed in the first column. We need to find the largest value in the data - value column.

Answer:

A. $A(t)=36100(1.05)^t$

13.