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sergij07 [2.7K]
3 years ago
11

A mixture contains forty ounces of glycol and water and is ten percent glycol. If the mixture is to be strengthened to twenty-fi

ve percent, how much glycol is to be added?
4 oz
8 oz
10 oz
Mathematics
2 answers:
irina1246 [14]3 years ago
7 0
10(40) + x = .25(x + 40)
 
4 + x = .25x + 10
 
x - .25x = 10 - 4
 
.75x = 6
x =
x = 8 oz of glycol required
telo118 [61]3 years ago
4 0
If we add 8 oz of glycol to the 40 oz mixture we have 48 oz of mixture and 4+8 = 12 oz glycol
 12 / 48  = 1/4   = 25 % glycol 

the answer is 8 oz
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Hi! 

To compare this two sets of data, you need to use a t-student test:

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t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

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5 0
3 years ago
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lisabon 2012 [21]

Given the following functions below,

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Factorising the denominators of both functions,

Factorising the denominator of f(x),

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Multiplying both functions,

undefined

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Answer:

$13

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