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Andrews [41]
2 years ago
8

The cost of having a package delivered has a base fee of $9.70. Every pound over 5 lbs cost an additional $0.46 per pound. Write

a linear model that represents the cost, C, for the total weight, w, in pounds of the package.
Mathematics
1 answer:
FrozenT [24]2 years ago
3 0
The linear model of this case takes the form:

y = a(x-b) + k

<span>The cost of having a package delivered has a base fee of $9.70
this is "k" >>>>>  k=9.7 (fixed amount of fee)

THEN

</span>
<span>Every pound over 5 lbs cost an additional $0.46 per pound

that means: 0.46(x-5)
in other words, if the package weighs foe example 9 pounds, then 9-5=4, it will cost 0.46*4 for these 4 extra pounds

Finally we have the linear form of this: C = 0.46 (W - 5) + 9.7
</span>
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Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

b > 3\frac{2}{15}\\\\

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality 2 \frac{3}{5} < b - \frac{8}{15}\\\\ is the same as \frac{13}{5} < b - \frac{8}{15}\\\\

---------------------

Let's multiply both sides by 15 to clear out the fractions

\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\

---------------------

Now isolate the variable b

39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

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