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Soloha48 [4]
2 years ago
6

Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $

1.25 for a delivery made within 1 1/2 mi, $1.70 for a delivery of at least 1 1/2 mi but less than 1 3/4 mi, $2.15 for a delivery of at least 1 3/4 mi but less than 2 miles, and so on. If tim raised his rates by 10% what would he be paid to deliver a package 3 1/8 miles
Mathematics
1 answer:
nlexa [21]2 years ago
6 0

Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

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7.24pm

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Answer:

<h2>length =  \frac{25}{28}</h2>

Step-by-step explanation:

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Substitute the values into the above formula and solve for the length

That's

<h3>length =  \frac{area}{width}</h3>

So we have

<h3>length =  \frac{25}{42}  \div  \frac{4}{6}  \\  =  \frac{25}{42}  \times  \frac{6}{4}  \\  =  \frac{25}{7}  \times  \frac{1}{4}</h3>

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