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miv72 [106K]
3 years ago
13

A car requires 22 litres of petrol to travel a distance of 259.6 km.Find

Mathematics
1 answer:
Anon25 [30]3 years ago
5 0

Answer:

Step-by-step explanation:

1. A car requires 22 litres of petrol to travel a distance of 259.6 km

what is the distance that the car can travel on 63 ltr of petrol

22ltr = 259.6km

63ltr=

cross multiply

{63 x 259.6}/22 = 16354.8/22 = 743.4 km

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 63 ltr of petrol to travel 743.4km

2. To travel a distance of 2013.2 km

we would need to calculate the amount of fuel

A car requires 22 litres of petrol to travel a distance of 259.6 km

what amount of fuel would it require to travel 2013.2km

22ltr = 259.6km

xltr = 2013.2km

x is the value of petrol to cover 2013.2km

cross multiply

(2013.2 x 22)/259.6

44290.4/259.6 = 170.610169492≈170.6 ltr

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 170.6 ltr of petrol to travel 2013.2km

if 1ltr is $1.99

170.6 ltr is (170.6 x 1.99)/1 = $339.494≈$339.5

The price of fuel consumed for 2013.2 km at 1 liter of petrol at $1.99 is $339.5

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General Idea:

To create a dot plot for a set of data we should list the data from lowest to greatest number. Then create a number line which covers all the numbers given in the data. We need to plot number of circles one above as many times based on the frequency of the data.

Applying the concept:

33, 36, 31, 37, 37, 38, 31, 37, 35, 31, 38, 32, 36, 33, 38

Listing the data in ascending order we get

31, 31, 31, 32, 33, 33, 35, 36, 36, 37, 37, 37, 38, 38, 38

We need to plot <u>three </u>circles above 31, <u>one</u> circle above 32, <u>two</u> circles above 33, <u>one</u> above 35, <u>two </u>above 36, <u>three </u>above 37, and <u>three</u> above 38.

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Out of the four options, the <u>correct option is B </u>because that is the only dot plot representing the given set of data.

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3 years ago
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If 9m=5n then m/n=?
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Step-by-step explanation:

9m = 5n

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7 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD=√2 cm.
Zielflug [23.3K]

Given that triangle ABC is right angle triangle. CD is altitude such that AD=BC

ABC is a right angle triangle so apply Pythogorean theorem

AC^{2} + BC^{2} = AB^{2}

AC^{2} + AD^{2} = AB^{2}      (Given that AD = BC)

AC^{2} + AD^{2} = 3^{2}      (Given that AB=3)

AC^{2} + AD^{2} = 9 ...(i)

ADC is a right angle triangle so apply Pythogorean theorem

AD^{2} + CD^{2} = AC^{2}

AD^{2}+(\sqrt{2})^2= AC^{2}

AD^{2}+2= AC^{2}

AD^{2}=AC^{2} -2 ...(ii)


Plug value (ii) into (i)

AC^{2} + AC^{2}-2 = 9

2AC^{2} -2=9

2AC^{2} =11

AC^{2} =\frac{11}{2}

AC=\sqrt{\frac{11}{2} }


Hence final answer is AC=\sqrt{\frac{11}{2} }

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<h3>Properties ofa rhombus</h3>

The given diagram is a rhombus that has the following properties

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Learn more on rhombus here: brainly.com/question/88523

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