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3 years ago

7

How many ounces of 10% alcohol would need to be mixed with pure rubbing alcohol to get 16

1 answer:

givi [52]3 years ago

6 0

Suppose you add <em>x</em> oz of 10% alcohol to <em>y</em> oz of pure alcohol. Then the mixture has a total volume of <em>x</em> + <em>y</em> oz, and we want to end up with 16 oz so that

<em>x</em> + <em>y</em> = 16

For each oz of the solution 10% used, 0.1 oz of alcohol is contributed, and each oz of pure alcohol contributes 1 oz of alcohol. The mixture is supposed to have a concentration of 14%, which comes out to 0.14*16 = 2.24 oz of alcohol. Then

0.1<em> x</em> + 1 <em>y</em> = 2.24

Solve for <em>y</em> in both equations:

<em>y</em> = 16 - <em>x</em>

<em>y</em> = 2.24 - 0.1 <em>x</em>

Set them equal to one another and solve for <em>x</em>, then for <em>y</em>.

16 - <em>x</em> = 2.24 - 0.1 <em>x</em>

13.76 = 0.9 <em>x</em>

<em>x</em> = 13.76/0.9 ≈ 15.29

<em>y</em> = 16 - 15.29 ≈ 0.71

So you need about 15.29 oz of 10% alcohol and 0.71 oz of pure alcohol to get the desired mixture.

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xeze [42]

Answer: 1. 0.0256

2. 0.4096

Step-by-step explanation:

Binomial probability formula , to find the probability of getting x successes:

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where n= Total number of trials

p= Probability of getting success in each trial.

Let x be the number of customers will make purchase.

As per given , we have

p= 0.20

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1. The probability that 3 of the next 4 customers will make a purchase will be:-

$P(x=3)=^4C_3(0.20)^3(1-0.20)^{4-3}$

$P(x=3)=(4)(0.20)^3(0.80)^{1}\ \ [\because\ ^nC_{n-1}=n]$

$P(x=3)=(4)(0.008)(0.80)=0.0256$

Hence, the probability that 3 of the next 4 customers will make a purchase = 0.0256

2. The probability that none of the next 4 customers will make a purchase will be :

$P(x=0)=^4C_0(0.20)^0(1-0.20)^{4-0}$

$P(x=0)=(1)(0.80)^{4}\ \ [\because\ ^nC_{0}=1]$

$P(x=0)=0.4096$

Hence, the probability that none of the next 4 customers will make a purchase= 0.4096

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