No, there’s a capacity of 4 students because 4(number of buses)* 30(the capacity of each bus) gives you 120 and there’s a total of 24 students because 24(#of kids in each class)* 6(# of classes) equals 144
The answer is: " 128 oz. " .
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There are: " 128 oz. " (in " 8 lbs." ) .
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Explanation:
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Set up a proportion; as a fraction; as follows:
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400/ 25 = x / 8 ;
in which: "x" = the number of "ounces [oz.] there are in "8 lbs." ;
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We shall solve for "x" , the answer to the problem:
Cross-factor multiply:
25x = (400) * 8 ;
→ 25x = 3200 ;
Divide each side of the equation by "25" ; to isolate "x" on one side of the equation; & to solve for "x" ;
→ 25x / 25 = 3200 / 25 ;
→ x = 128 .
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Answer: " 128 oz. " .
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There are: " 128 oz. " (in " 8 lbs." ) .
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Note of interest: " 16 oz. = 1 lb. " (exact conversion).
So; "8 lbs. <span>= ?</span> oz. " ;
→ " 8 lbs. * (16 oz/ 1 lb) = ( 8 * 16) oz. = 128 oz. ; → which is our answer!
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Answer: 98268
Step-by-step explanation:
Given: In an international company,
Employees in one country = 22,700
If this is 23.1% of the total number of company's employees,
[
, replace 'of' by '×']
i.e. 23.1% of (total employees) = 22700
![\Rightarrow\ \dfrac{23.1}{100}\times\text{(total employees)}=22700\\\\\Rightarrow\ \text{Total employees}=\dfrac{22700\times100}{23.1}\\\\\Rightarrow\ \text{Total employees}=98268.4\approx98268](https://tex.z-dn.net/?f=%5CRightarrow%5C%20%5Cdfrac%7B23.1%7D%7B100%7D%5Ctimes%5Ctext%7B%28total%20employees%29%7D%3D22700%5C%5C%5C%5C%5CRightarrow%5C%20%5Ctext%7BTotal%20employees%7D%3D%5Cdfrac%7B22700%5Ctimes100%7D%7B23.1%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Ctext%7BTotal%20employees%7D%3D98268.4%5Capprox98268)
Hence, there are 98268 employees in total.
Answer:
10 the next day
Step-by-step explanation:
Using the least common multiple
3, 6, 8
The least common multiple is 24
24 hours from 10 the next day
The area of a circle is given by
![A = \pi r^2](https://tex.z-dn.net/?f=%20A%20%3D%20%5Cpi%20r%5E2%20)
whereas the circumference is given by
![C = 2\pi r](https://tex.z-dn.net/?f=%20C%20%3D%202%5Cpi%20r%20)
If we want these two values to be (numerically) the same we have to set
and solve for the radius:
![A = C \iff \pi r^2 = 2\pi r \iff r^2 = 2r \iff r^2-2r=0 \iff r(r-2) = 0](https://tex.z-dn.net/?f=%20A%20%3D%20C%20%5Ciff%20%5Cpi%20r%5E2%20%3D%202%5Cpi%20r%20%5Ciff%20r%5E2%20%3D%202r%20%5Ciff%20r%5E2-2r%3D0%20%5Ciff%20r%28r-2%29%20%3D%200)
So, one (trivial) solution is
. A circle with radius 0 is just a point, and so both area and circumference are zero.
The other solution is
. In fact, you have
![A = \pi r^2 = 4\pi,\quad C = 2\pi r = 2\pi \cdot 2 = 4\pi](https://tex.z-dn.net/?f=%20A%20%3D%20%5Cpi%20r%5E2%20%3D%204%5Cpi%2C%5Cquad%20C%20%3D%202%5Cpi%20r%20%3D%202%5Cpi%20%5Ccdot%202%20%3D%204%5Cpi%20)