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2 years ago

11

Carlos has 7 1/2 acres of farmland. He uses 1/3 of the acres to graze animals and 1/5 of the acres to grow vegetable. How many a

cres does Carlos use for grazing animals or for growing vegetables?

1 answer:

omeli [17]2 years ago

3 0

8/15 or somthing like that it you might be souposed to solve it diferent than i did

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If lmn xyz which congruences are true cpctc check all that apply i.will give you brainiest

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2 years ago

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How many natural numbers are there between 8² and 9²

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Soccer game theory is that one of the first people to play the game and play it they will

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2 years ago

A stadium brings in $16.25 million per year. it pays football-related expenses of $13.5 million and stadium expenses of $2.7 mil

Ganezh [65]

We have that

Profit margin is calculated by finding the net profit as a percentage of the revenue

Profit margin = [Net profit / Revenue]

Net profit= [Revenue-Cost]

we know that

Revenue = $16.25 million

Cost = $13.5 million + $2.7 million

Net profit = [16.25 million - (13.5 million + 2.7 million)]

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Profit margin = 0.05 / 16.25

Profit margin = 0.003077 or 0.3077%

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2 years ago

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

nlexa [21]

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%

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

6 0

2 years ago

Mindy's dog, Rex, weighed 80 pounds a year ago. Now he weighs 92 pounds. What is the percent increase in his weight?

nadya68 [22]

Answer:

15%

Step-by-step explanation:

To find the percent increase, use the equation:

percent of increase=<u>amount of increase</u>

original amount

The amount of increase is 92-80 = 12. The original amount of 80; this gives us

12/80 = 0.15 = 15%

5 0

2 years ago