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hram777 [196]
3 years ago
14

12 out of 18 students buy pizza for lunch on Fridays. At this rate if there are 240 students in the entire school, how many stud

ents can be expected to buy pizza for lunch this Friday?
Mathematics
1 answer:
laiz [17]3 years ago
3 0
160. This is because u take the total(240) divide it by 18 and multiply by 12.


however, you can also simplifiy the fraction (12/18) into (2/3) and divide by the bottom denominator (3) and multiply by the top numerator. ;)
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24 is 40% of what number?
Murrr4er [49]
24=\frac{40}{100}*x \\ x=24*\frac{5}{2}=60
24 is 40% of 60.
5 0
3 years ago
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An in ground rectangular pool has a concrete pathway surrounding the pool. if the pool is 16 feet by 32 feet and the entire area
Anastasy [175]
1. Let's call:
 
 x: the width of the walkway.
 b: 2x+16 (there is a widht "x" of the walkway on each side).
 h: 2x+32  (there is a widht "x" of the walkway on each side).
 A: 924 ft^2 ( the area of the pool of including the walkway).
 
 2. The area of a rectangle is:  
 
 A=(b)(h)
 
 b: the base of the rectangle (2x+16)
 a: the height of the rectangle (2x+32)

 3. Then, we have a quadratic equation:
 
 924=(2x+16)(2x+32)
 4x^2+64x+32x+512-924=0
 4x^2+96x-412=0
 
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 The answer is: The width of the walkway is 3.71 ft

7 0
3 years ago
HELP PLZZZZZ !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Neporo4naja [7]

Answer: I think it’s  A because 8x5=40

Step-by-step explanation:

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3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
Someone please help ASAP. I would really really appreciate it if u can. I will mark brainliest
AlladinOne [14]

Answer:

y = -\frac{1}{4} x + 2

Step-by-step explanation:

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