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Jillane babysat her younger sister for [tex]$2 \frac{1}{2}$[/tex] hours on Friday and [tex]$\frac{2}{3}$[/tex] hour on Saturday. How much longer did she babysit on Friday than on Saturday?

Sagot :

GinnyAnsweridn
To determine how much longer Jillane babysat her younger sister on Friday compared to Saturday, we can break down the problem into steps:

1. Convert the mixed number to an improper fraction or decimal:
- On Friday, Jillane babysat for [tex]\(2 \frac{1}{2}\)[/tex] hours. In decimal form, [tex]\(2 \frac{1}{2}\)[/tex] hours is equivalent to 2.5 hours.

2. Convert the fraction to a decimal:
- On Saturday, Jillane babysat for [tex]\(\frac{2}{3}\)[/tex] of an hour. In decimal form, [tex]\(\frac{2}{3}\)[/tex] is approximately 0.6667 hours.

3. Compare the two amounts to find the difference:
- To determine how much longer she babysat on Friday, we subtract the hours babysat on Saturday from the hours babysat on Friday.

So we calculate:

[tex]\[ 2.5 \, \text{hours (Friday)} - 0.6667 \, \text{hours (Saturday)} = 1.8333 \, \text{hours} \][/tex]

Therefore, Jillane babysat her younger sister 1.8333 hours longer on Friday than on Saturday.
Chibabykalu16idn

Answer:

She spent 1 5/6 hours longer on Friday

Step-by-step explanation:

Given:

  • Jillane babysat her younger sister for [tex] 2 \frac{1}{2} [/tex] hours on Friday
  • Jillane babysat her younger sister for [tex] \frac{2}{3} [/tex] hours on Saturday

To compare the duration and find the difference, we need to convert the mixed number [tex] 2 \frac{1}{2} [/tex] into an improper fraction for easier comparison.

[tex]2 \frac{1}{2} = \frac{(2\times2)+1}{2} = \frac{5}{2}[/tex]

Solving for the difference...

5/2 - 2/3

Find the L.C.M (Lowest Common multiple) of 2 and 3 which are the denominators which is 6

[tex]\Large \frac{(\frac{5}{2} \times 6) - (\frac{2}{3} \times 6)}{6}[/tex]

(15 - 4) / 6

= 11/6 or  [tex]1 \frac{5}{6}[/tex] hours

Therefore, the final answer is:

[tex]\Large \boxed{\boxed{1 \frac{5}{6} \approx 1.83 \: (\text{hours})}}[/tex]

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