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2 years ago

Sandeep used 8 litres of fuel to travel 76 miles.

Mathematics

1 answer:

OleMash [197]2 years ago

4 0

Answer:

I got, when calculated, 209 miles. (It may be wrong)

Step-by-step explanation:

First, I calculated how many miles can Sandeep travel with 1L, therefore I divided 76 miles by 8L and got 9.5 miles on 1L.

Then I multiplied 9.2 miles by 22L and got 209 miles.

This may be wrong but if anyone else is able to work it out much appreciated if it's correct.. Your most welcome!

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Round 8.78911981582 to the nearest millionth

Aliun [14]

Answer:

8.789120

(you have to include the zero, or else the answer would no longer be the nearest millionth! and you would be marked wrong)

8 0

2 years ago

Here are yesterday's high temperatures (in Fahrenheit) in 12 U.S. cities. 48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80 Notice

irinina [24]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

Learn more on Measures of a Data here: brainly.com/question/15097997

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1 year ago

Can someone help me with this geometry question?

lana66690 [7]

Answer:

whats the question

Step-by-step explanation:

3 0

2 years ago

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Factorize (x-3)(x-5)(x-7)(x-9)+15

zhannawk [14.2K]

The expanded form of the equation is x^4-24x^3+222x^2-240+960

<h3>Expansion of expression</h3>

Given the expression below;

(x-3)(x-5)(x-7)(x-9)+15

Expand

(x-3)(x-5)(x-7)(x-9) + 15

(x^2-5x-3x+15)(x^2-9x-7x+63) +15

(x^2-8x+15)(x^2-16x+63) + 15

Expand further to have;

x^4-16x^3+63x^2-8x^3+144x^2-504+15x^2-240x+945+15

x^4-24x^3+222x^2-240+960

Hence the expanded form of the equation is x^4-24x^3+222x^2-240+960

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6 0

1 year ago

Hello,

Nookie1986 [14]

Answer:

s = height above ground

s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)

at t = 0

s = 60 (clearly :)

now when does it hit the bleak earth?

That is when s = 0

4.9 t^2 - 20 t -60 = 0

solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)

t = 6.09 or - 2.01

use t = 6.09

now to do the last part there are two obvious ways to get t at the peak

1. look for vertex of parabola

2. look for halfway between t = -2.01 and t = 6.09

I will do it the hard (11) waay by completing the square

4.9 t^2 - 20 t = -(s-60)

t^2 - 4.08 t = -.204 s + 12.2

t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16

(t-2.04)^2 = -.204(s-80.2)

top at 80.2 meters at t = 2.04 s

===============

quick check on time

should be average of 6.09 and -2.01

=4.08 /2 = 2.04 check

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1 year ago

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