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AysviL [449]
2 years ago
12

Saline solution A is 10% strength and saline solution B is 5% strength. I have an order for 6 litres of an 8% saline solution. T

o fill the order how many litres of solution A should be diluted with solution B? (Hint. Assume a 1% solution contains r grams of salt per litre.)
Mathematics
2 answers:
kupik [55]2 years ago
7 0
Let x = volume of saline solution A,
y = volume of saline solution B.

We're going to set up two equations. 
Firstly, x + y = 6. That gives us y = 6 – x     (1)

Second, 0.1x + 0.05y = 0.08•6, 
or 0.1x + 0.05y = 0.48               (2)

Substitute (1) into (2):
0.1x + 0.05(6 – x) = 0.48
0.1x + 0.3 – 0.05x = 0.48
0.05x + 0.3 = 0.48

Subtract 0.3 from both sides:
0.05x = 0.18
x = 3.6
Then y = 6 – 3.6 = 2.4

So we have to use 3.6 litres of solution A and 2.4 litres of solution B.


morpeh [17]2 years ago
5 0
<span>18/5 (or 3.6) liters of solution A should be diluted with solution B.</span>
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2 years ago
A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

                Z = \frac{X-\mu}{\sigma} ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

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3 years ago
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