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lana66690 [7]
2 years ago
12

write and solve a real- world problem involving the multiplication of a fraction and a whole number whose product is between 10

and 15
Mathematics
2 answers:
Amanda [17]2 years ago
7 0
Tom has a bag of 30 oranges. He says his friend, Ann can have half. How many oranges do they each get? 1/2 x 30 = 15
Tomtit [17]2 years ago
4 0

Answer:

Real word Problem:- Jean is working in a pizza shop. She has 60 kg of flour. She has used one-fifth of it for making pizzas. How much flour she used for making pizzas?

Solution:- Quantity of flour she has= 60 kg

The quantity she used in making pizzas=\frac{1}{5}\ of\ 60

=\frac{1}{5}\times60=12\ kg

hence, she used 12 kg of flour for making pizzas.

Here the multiplication of a fraction and a whole number =\frac{1}{5}\times60 and its answer is between 10 and 15, since 10<12<15.

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Help guys with this question
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the answer is 1/2!!

Step-by-step explanation:

Good luck at that quiz izz!

Hope this helps! :)

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Number 49....I need help
sergiy2304 [10]
Now, keep in mind that, we'll be using 25mile intervals, notice, that at a cost of 200 bucks, it went for 50 miles, now the charges are just for every 25miles only, so, if you drive 30 miles, you pay for 25miles and if you drive for 49 miles, you still only pay for 25 miles, but because 50/25 is 2 intervals, then if you drive for 50 miles, you're paying for 2 intervals of 25miles, not just one.

now, also let's keep in mind that, when the car was driven for 200 miles, that's 200/25 or 8 25mile intervals, so the 25mile charge kicked in 8 times.

\bf \stackrel{cost}{y}=\stackrel{fixed~charge}{F}+\stackrel{25mile~intervals}{m}\quad \stackrel{25mile~charge}{x}\implies y=F+mx\\\\&#10;-------------------------------\\\\&#10;\begin{cases}&#10;200=F+2x\implies 200-2x=\boxed{F}\\&#10;245=F+8x\\&#10;----------\\&#10;245=\boxed{200-2x}+8x&#10;\end{cases}&#10;\\\\\\&#10;45=6x\implies \cfrac{45}{6}=x\implies \cfrac{15}{2}=x\impliedby &#10;\begin{array}{llll}&#10;\textit{so the 25mile charge is}\\&#10;\textit{7 bucks and 50cents}&#10;\end{array}&#10;\\\\\\&#10;

\bf 200=F+2\left( \frac{15}{2} \right)\implies 200=F+15\implies 185=F&#10;\\\\\\&#10;\textit{and the \underline{fixed charge} is 185 bucks then, let's check for 450}&#10;\\\\\\&#10;\boxed{y=185+\frac{15}{2}x}\qquad &#10;\begin{cases}&#10;450~miles\\\\&#10;\cfrac{450}{25}\implies 18~\textit{\underline{25mile intervals}}\\\\&#10;x=18&#10;\end{cases}&#10;\\\\\\&#10;y=185+\cfrac{15}{2}\cdot 18\implies y=185+135\implies y=320
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3 years ago
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