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ozzi
3 years ago
14

A pharmacist has 40% and 60iodine solutions on hand. How many liters of each iodine solutions will be required to produce 4 lite

rs of a 50% iodine mixture
Mathematics
1 answer:
Schach [20]3 years ago
4 0

A pharmacist has 40% and 80% of iodine solutions on hand. How many liters of each iodine solution will be required to produce 4 liters of a 50% iodine mixture?

.

Let x = liters of 40% iodine

then

4-x = liters of 80% iodine

Using algebra:

.40x + .80(4-x) = .50(4)

.40x + 3.20-.80x = 2

3.20-.40x = 2

x = 4 liters (40% iodine)

80% iodine:

4-x = 4-4 = 0 liters needed (80% iodine)

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Describe the relationship between two variables when the correlation coefficient r is one of the following. (a) near –1 strong
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Now back to the question:

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3 years ago
An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed
masya89 [10]

Answer:

P(X\leq 1) = 0.331

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

P(X = x) = \frac{y^xe^{-y}}{x!}

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

P(X\leq 1) = P(X=0) + P(X =1)

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}

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P(X = 0) = 0.10025884372

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}

P(X = 1) = \frac{2.3 * e^{-2.3}}{1}

P(X = 1) =2.3 * e^{-2.3}

P(X = 1) = 2.3 * 0.10025884372

P(X = 1) = 0.23059534055

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P(X\leq 1) = 0.10025884372 + 0.23059534055

P(X\leq 1) = 0.33085418427

P(X\leq 1) = 0.331

Hence, the required probability is 0.331

6 0
2 years ago
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