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Nataliya [291]
4 years ago
9

rainer bought 1.43 pounds of lunch meat at the store for 10.98.what was the unit rate (price) of the lunch meat?

Mathematics
2 answers:
Fed [463]4 years ago
5 0

Answer : The unit rate (price) of the lunch meat was, $7.68

Step-by-step explanation :

As we are given that the rainer bought 1.43 pounds of lunch meat at the store for $10.98.

Now we have to determine the unit rate (price) of the lunch meat.

As, the rate of lunch meat of 1.43 pounds = $10.98

So, the rate of lunch meat of 1 pounds = \frac{1\text{ pounds}}{1.43\text{ pounds}}\times \$ 10.98

                                                                = $7.68

Thus, the unit rate (price) of the lunch meat was, $7.68

omeli [17]4 years ago
3 0
If I did the math correctly, it should be $7.86 per pound.
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3 years ago
What is the difference in speed between two cars when one traveled 175 km in 4 hours and the other traveled the same distance in
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<h3>Given :</h3>
  • Distance = 175
  • Time 1 = 4 hours

\\  \\

<h3>To find:</h3>
  • Different between two speeds when distance is same.

\\  \\

<h3>Solution:</h3>

<u>Part</u><u> </u><u>1</u><u> </u><u>:</u>

In first part we will find speed 1 in which time is 4 hours

we know:-

\bigstar \boxed{ \rm speed =  \frac{distance}{time} }

So:-

\dashrightarrow\sf speed_1 =  \dfrac{distance}{time_1}

\\  \\

\dashrightarrow\sf speed_1 =  \dfrac{175}{4}

\\  \\

\dashrightarrow\bf speed_1 =  43.75 \: km {h}^{ - 1}

<u>Part 2</u>

In first part we will find speed 2 in which time is 3 hours

Again:

\bigstar \boxed{ \rm speed =  \frac{distance}{time} }

\\  \\

\dashrightarrow\sf speed_2 =  \dfrac{distance}{time_2}

\\  \\

\dashrightarrow\sf speed_2 =  \dfrac{175}{3}

\\  \\

\dashrightarrow\bf speed_2 = 58.33kmh^{-1}

\\  \\

<u>Part 3</u>

In this part we will find different between speed 1 and speed 2

\bigstar  \boxed{\rm Difference = speed_2 - speed_1}

\\  \\

So :

\dashrightarrow\sf Difference = speed_2 - speed_1 \\

\\  \\

\dashrightarrow\sf Difference =58.33 - 43.75\\

\\  \\

\dashrightarrow\bf Difference =14.58kmh^{-1}\\

\\  \\

\therefore \underline {\textsf{\textbf{Difference in speed between two cars is \red{14.58km/h}}}}

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3 years ago
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