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FrozenT [24]
2 years ago
10

A candy manufacturer makes two types of special candy, say A and B. Candy A consists of equal parts of dark chocolate and carame

l and Candy B consists of two parts of dark chocolate and one part of walnut. The company has in stock 430 kilograms of caramel, 360 kilograms of dark chocolate, and 210 kilograms of walnuts. The company sells Candy A for P285 and Candy B for P260 per kilograms. How much of each candy should the manufacturer produce to maximize profit?
Mathematics
1 answer:
Sphinxa [80]2 years ago
6 0
If we let x as candy A
y as candy B
a as dark chocolate in candy a
b as dark chocolate in candy b
c as caramel
d as walnut
P as profit
we have the equations:
a + c = x
2b + d = y
a + 2b ≤ 360
c ≤ 430
d ≤ 210
P = 285x + 260y

This is an optimization problem which involves linear programming. It can be solved by graphical method or by algebraic solution.
P = 285(a + c) + 260(2b +d)
If we assume a = b
Then a = 120, 2b = 240
P = 285(120 + 120) + 260(240 + 120)
P = 162000
candy A should be = 240
candy B should be = 360
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2 years ago
Consider the lengths of stay at a hospital’s emergency department. Hours Count Percent 1 18 3.44 2 55 10.50 3 81 15.46 4 109 20.
Vladimir [108]

Answer:

Probability = 0.502

Step-by-step explanation:

We are given the following data :

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  1                    18               3.44

  2                    55              10.50

  3                    81               15.46

  4                    109             20.80

  5                     88              16.79

  6                     66              12.60

  7                     39               7.44

  8                     17                3.24

  9                     17                3.24

  10                   19                3.63

   15                  15                2.86

We need to calculate the probability

P(Length of stay of exactly 1 is less than or equal to 4)

P(Y \leq 4) = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

P(Y \leq 4) = 0.0344 + 0.1050 + 0.1546 + 0.2080 = 0.502

We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.

8 0
2 years ago
A test tube contains 25 bacteria, 5 of which are can stay alive for atleast 30 days, 10 of which will die in their second day. 1
boyakko [2]

Answer:

c. \frac{1}{5}

Step-by-step explanation:

Given,

Number of bacterias who alive for at least 30 days = 5,

Bacterias who alive for 2 days = 10,

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Total bacterias = 5 + 10 + 10 = 25,

Ways of choosing a bacteria = ^{25}C_1 = \frac{25!}{1! 24!} = 25,

While, ways of choosing of a bacteria who will live after 1 week = ^5C_1 = \frac{5!}{1!4!} = 5,

Hence, the probability it will still be alive after one week = \frac{5}{25} = \frac{1}{5}

OPTION C is correct.  

3 0
2 years ago
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