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2 years ago

Larry mixed 15 grams of salt with 210 grams of a 5% salt solution in water. What is the percent of salt in his new solution?

1 answer:

5 0

When the concentration of a solution is expressed in weight percentages, that is just equal to the mass of solute over the mass of the total solution. For a solution weighing 210 grams, 5% of it is salt. That is equal to

210*0.05= 10.5 g salt

The rest of the amount, determined by difference, is the amount of solvent which is water.

water = 210-10.5 = 199.5 g salt

To solve for the new concentration, you add up 15 g to the already existing 10.5 g of salt in the solution. The water, on the other hand, remains constant.

Concentration = (15+10.5)/(210+15) = 0.1133 or 11.33%

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Hmm... well, here ya go.

1.50 ÷ 0.05 = 30
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soldier1979 [14.2K]

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2 years ago

Which function has the domain x>or= -11

iogann1982 [59]

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2 years ago

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Nadusha1986 [10]

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Step-by-step explanation:

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2 years ago

let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo

anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

$f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\$

a) P(x > 5) = $\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865$

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = $\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}$

But P(x > a) = 0.05

Therefore:

$e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98$

a ≅ 12

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2 years ago