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Anestetic [448]
2 years ago
12

On average, a commercial airplane burns about 3.9 × 10³ mL of gasoline per second. There can be as many as 5.1 × 10³ airplanes i

n the air in the United States at any given time.
About how many milliliters of gasoline are burned per minute by commercial airplanes in flight in the United States?

Enter your answer in the box.
1.2 x 10 to what power?
Mathematics
2 answers:
algol132 years ago
8 0
We use the given data above to calculate the volume of gasoline that is being burned per minute by commercial airplanes. 

Amount burned of 1 commercial airplane = <span>3.9 × 10³ ml of gasoline per second
Number of airplanes = </span><span>5.1 × 10³ airplanes

We calculate as follows:

</span> 3.9 × 10³ ml of gasoline per second / 1 airplane (5.1 × 10³ airplanes)(60 second / 1 min ) = <span>1.2 x 10^9 mL / min</span>
lyudmila [28]2 years ago
5 0

Answer: 1.2x10^9 mL/min

Step-by-step explanation:

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Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate
mars1129 [50]

Answer: In the beginning he was given 27 sweets.

Step-by-step explanation: The most logical thing to do is to solve it backwards, that is, from what he had at the end of the third day up till the beginning of the first day.

On the third day he ate one-third and had 8 sweets left over. To determine how many he started with on the third day, let the total on day three be called a. If one-third of a is eaten, then the left over which is two-thirds is 8. That is;

8/a = 2/3

By cross multiplication we now have

8 x 3 = 2a

24/2 = a

a = 12

Let the number of sweets he had on day two be called b. If he ate one-third of b and he had 12 left over, then the two-thirds left over is 12 and we now have;

12/b = 2/3

By cross multiplication we now have

12 x 3 = 2b

36 = 2b

36/2 = b

b = 18

Let the number of sweets he had on day one be called x. If he ate one-third of x and he had 18 left over, then the two-thirds left over is 18, and we now have;

18/x = 2/3

By cross multiplication we now have

18 x 3 = 2x

54 = 2x

x = 27

Therefore Tim was given 27 sweets at the beginning.

3 0
2 years ago
During a certain week, a post office sold Rs.280 worth of 14-paisas stamps. How many of these stamps did they sell?
Novosadov [1.4K]
So basically ...

You convert the rupees in paisas. One rupee is equal to one hundred paisas, so ...

280 × 100 = 28,000

And then we divide,

28,000 ÷ 14 = 2000

The post office sold 2000 stamps!

Hope this helped! :)
6 0
2 years ago
Jessica rides the bus 8and 4/5 miles each day. WHATS is The total Numbers of miles she rindes in 21 day's
lesantik [10]
201 miles i believe, hope this helps!
5 0
2 years ago
Read 2 more answers
The data set below represents the ages of 36 executives. find the percentile that corresponds to an age of 4141 years old. 2828
solong [7]

Answer:

37th percentile.

Step-by-step explanation:

We have been given a data set that represents the ages of 36 executives. We are asked to find the percentile that corresponds to an age of 41 years.

28, 29, 29, 32, 32, 33, 34, 34, 34, 34, 37, 37, 38, 41, 41, 42, 45, 45, 47, 47, 47, 48, 50, 51, 53, 56, 56, 56, 61, 61, 62, 63, 64, 64, 65, 66.

Let us count the number of data points below and at 41.

We can see that the number of data points at and below 41 is 13.

We will use percentile formula to solve our given problem.

\text{Percentile rank of x}=\frac{\text{Number of values below x}}{\text{Total number of data points}}\times 100

\text{Percentile rank of 41}=\frac{13}{36}\times 100

\text{Percentile rank of 41}=0.361111\times 100

\text{Percentile rank of 41}=36.11\approx 37

Therefore, the percentile rank that corresponds to age of 41 years old is 37th percentile.  

8 0
2 years ago
se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y
alexandr1967 [171]

Answer:

It is proved that f_x, f_y exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)

  • To show exixtance of f_x(0,0), f_y(0,0) we take,

f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

And,

f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}   exists.

  • To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0
2 years ago
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