# Question #1ca5a

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

These are classic examples of **unit conversion** problems that can be solved by using one or more *conversion factors* that help you go from one unit to another.

As far as I know, the most common conversion factor used to convert between *milliliters* to *drops* and vice versa is

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 mL " = " 20 drops")color(white)(a/a)|)))#

This tells you that in order to have **drops** of soda. Since you know that a can of soda contains *drops* would be needed to get that volume

#355 color(red)(cancel(color(black)("mL soda"))) * "20 drops"/(1color(red)(cancel(color(black)("mL soda")))) = color(green)(|bar(ul(color(white)(a/a)"7100 drops"color(white)(a/a)|)))#

The exact same approach can be used to determine how many bales of hay will be consumed in **one year**. Unless the problem says otherwise, you can usually *approximate* one year to be equivalent to **weeks**

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 year " = " 52 weeks")color(white)(a/a)|)))#

This time, you know that the herd consumes **weeks**, which is equivalent to saying that they consume

#1 color(red)(cancel(color(black)("week"))) * "14 bales of hay"/(2color(red)(cancel(color(black)("week")))) = "7 bales of hay"#

in **one week**. Since you need **times** more hay in one year than it does in one week.

#7color(white)(a) "bales of hay"/(1color(red)(cancel(color(black)("week")))) * (52color(red)(cancel(color(black)("weeks"))))/"1 year" = color(green)(|bar(ul(color(white)(a/a)"364 bales of hay/year"color(white)(a/a)|)))#