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2 years ago

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Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req

uested, they will dry the casein to 12% moisture. The drying costs are $5/100 kg of water removed. Assume that that the dairy has 1000 kg of the original product (wet casein) that will be dried to 12% moisture and then sold. How much dried casein is produced (in kg)?
The answer is 852 (with margin 1). Why is it?

1 answer:

Nataly_w [17]2 years ago

3 0

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

Learn more about mass at: brainly.com/question/24658038

#SPJ1

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= 8480/69

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∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

=7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

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= 1104600/69 - (122.9)²

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Var = 904.2856

S.D = √Var

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σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

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Using the normal distribution probability table, we can compute the value of P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

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P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

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μ = 115 - 75.93

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3 years ago