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2 years ago

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Ms. Martinez has a class of 16 students. She can spend $17 on each student to buy math supplies for the year. She first buys all

of her students calculators,
which costs a total of $89.76. After buying the calculators, how much does she have left to spend on each student?
su

1 answer:

amid [387]2 years ago

3 0

Answer:

none

Step-by-step explanation:

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What conclusions can be made about the amount of money in each account if f represents Molly's account and g represents her brot

Nat2105 [25]

Answer:

(b) is true

Step-by-step explanation:

Given

Molly

$a = 500$ --- starting balance

$m = 10$ --- monthly rate

Her brother

$a = 100$ ---- starting balance

$r = 10\%$ --- annual rate

Required

Determine which option is true

First, we calculate her brother's function.

The function is an exponential function calculated as:

$y = ab^x$

Where $b = 1 + r$

So, we have:

$y = ab^x$

$y = 100 *(1 + 10\%) ^x$

$y = 100 *(1 + 0.10) ^x$

$y = 100 *(1.10) ^x$

Hence:

$g(x) = 100 *(1.10) ^x$

Next, we calculate Molly's function (a linear function)

The monthly function is:

$y = mx + a$

So, we have:

$y = 10x + 500$

Annually, the function will be:

$y = 10x*12 + 500$

$y = 120x + 500$

So, we have:

$f(x) = 120x + 500$

At this point, we have:

$f(x) = 120x + 500$ ---- Molly

$g(x) = 100 *(1.10) ^x$ ---- Her brother

<u>Next, we test each option</u>

(a): Molly's account will have a faster rate of change over [32,40]

We calculated Molly's function to be:

$y = 120x + 500$

The slope of a linear function with the form: $y = mx + b$ is m

By comparison:

$m = 120$

Since Molly's account is a linear function, the rate of change over any interval will always be the same; i.e.

$m = 120$

For his brother:

Rate of change is calculated using:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(40) - g(32)}{40 - 32}$

$m = \frac{g(40) - g(32)}{8}$

Calculate g(40) and g(32)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{40} =4526$

$g(32) = 100 * 1.10^{32} = 2111$

So, we have:

$m = \frac{4526 - 2111}{8}$

$m = \frac{2415}{8}$

$m = 302$

By comparison: $302 > 120$

Hence, her brother's account has a faster rate over [32,40]

(a) is false

(b): Molly's account will have a slower rate of change over [24,30]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(30) - g(24)}{30 - 24}$

$m = \frac{g(30) - g(24)}{6}$

Calculate g(30) and g(24)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{30} =1745$

$g(32) = 100 * 1.10^{24} = 985$

So, we have:

$m = \frac{g(30) - g(24)}{6}$

$m = \frac{1745 - 985}{6}$

$m = \frac{760}{6}$

$m = 127$

By comparison: $127 > 120$

Hence, Molly's account has a slower rate over [24,30]

(b) is false

(c): Molly's account will have a slower rate of change over [0,4]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(4) - g(0)}{4 - 0}$

$m = \frac{g(4) - g(0)}{4}$

Calculate g(4) and g(0)

$g(x) = 100 *(1.10) ^x$

$g(4) = 100 * 1.10^4 =146$

$g(0) = 100 * 1.10^{0} = 100$

So, we have:

$m = \frac{g(4) - g(0)}{4}$

$m = \frac{146 - 100}{4}$

$m = \frac{46}{4}$

$m = 11.5$

By comparison: $120>11.5$

Hence, Molly's account has a faster rate over [0,4]

(c) is false

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