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3 years ago

Michael threw for 1,654 yards in his first five games. At this rate, how many yards will he have thrown for in fifteen games? ✗

Mathematics

2 answers:

vodka [1.7K]3 years ago

7 0

Answer:

4,962 yards.

Step-by-step explanation:

We are given that,

For the first 5 games, Michael threw 1,654 yards.

Since, we see that, the relationship between the yards and the games are directly proportional.

Then for 15 games,

Michael will have thrown for $\frac{1654}{5}\times 3$ yards i.e. 1654 × 3 = 4,962 yards

Hence, for 15 games, Michael have thrown for 4,962 yards

suter [353]3 years ago

6 0

Answer: 4,962 yards

Step-by-step explanation:

Given: Michael threw for 1,654 yards in his first five games.

The rate of threw = $\dfrac{\text{Distance}}{\text{Time}}$

Then , The rate of threw = $\dfrac{1654}{\text{5}}$ yard/ minute.

i.e. the distance of yards thrown in 1 minute = $\dfrac{1654}{\text{5}}$ yard

If the rate remains same , then the distance of yards thrown for in fifteen games is given by :-

$\dfrac{1654}{5}\times15=4,962\text{ yards}$

Hence, At this rate, 4,962 yards will he have thrown for in fifteen games.

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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)

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$y = mx + a$

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:

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

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

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By comparison: $302 > 120$

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

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(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)

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

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

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

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

$m = 11.5$

By comparison: $120>11.5$

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

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