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gladu [14]
3 years ago
8

You manage an ice cream factory that makes three flavors: Creamy Vanilla, Continental Mocha, and Succulent Strawberry. Into each

batch of Creamy Vanilla go 2 eggs, I cup of milk, and 2 cups of cream. Into each batch of Continental Mocha go 1 egg, 1 cup of milk, and 2 cups of cream, while into each batch of Succulent Strawberry go 1 egg, 2 cups of milk, and 1 cup of cream. You have in stock 350 eggs, 350 cups of milk, and 400 cups of cream. How many batches of each flavor should you make in order to use up all of your ingredients
Mathematics
1 answer:
Leokris [45]3 years ago
7 0

Answer:

<em>We must make 100 batches of Creamy Vanilla, 50 batches of Continental Mocha, and 100  batches of Succulent Strawberry</em>

Step-by-step explanation:

<u>System of Equations</u>

The problem will be modeled as a system of three equations with 3 unknowns. Let's call the following variables

x=Number of batches of Creamy Vanilla ice creams

y=Number of batches of Continental Mocha ice creams

z=Number of batches of Succulent Strawberry ice creams

We know there are 350 eggs available and each Creamy Vanilla uses 2 eggs, each Continental Mocha uses 1 egg, and each Succulent Strawberry uses 1 egg. This condition leads to the equation:

(1)  2x+y+z=350

Following the same reasoning, we set up the equation for the cups of milk

(2)  x+y+2z=350

Finally, the ingredients for the Succulent Strawberry leads to the last equation

(3)  2x+2y+z=400

The set of equations will be solved by the reduction method. Subtracting the equation (3) from the equation (1) we get

y=50

Multiplying the equation (2) by -2 and adding to the equation (1)

-y-3z=-350

solving for z

-3z=-350+y=-350+50

z=100

From equation (2)

x=350-y-2z=350-50-200

x=100

Thus, we must make 100 batches of Creamy Vanilla, 50 batches of Continental Mocha, and 100  batches of Succulent Strawberry

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Answer:

The second choice: Approximately 65.2\% of the pretzel bags here will contain between 225 and 245 pretzels.

Step-by-step explanation:

This explanation uses a z-score table where each z entry has two decimal places.

Let \mu represent the mean of a normal distribution of variable X. Let \sigma be the standard deviation of the distribution. The z-score for the observation x would be:

\displaystyle z = \frac{x - \mu}{\sigma}.

In this question,

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  • \sigma = 9.3.

Calculate the z-score for x_1 = 225 and x_2 = 245. Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.

\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}.

\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}.

The question is asking for the probability P(225 \le X \le 245) (where X is between two values.) In this case, that's the same as P(-1.61 \le Z \le 0.54).

Keep in mind that the probabilities on many z-table correspond to probability of P(Z \le z) (where Z is no greater than one value.) Therefore, apply the identity P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1) to rewrite P(-1.61 \le Z \le 0.54) as the difference between two probabilities:

P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61).

Look up the z-table for P(Z \le 0.54) and P(Z \le -1.61):

  • P(Z \le 0.54)\approx 0.70540.
  • P(Z \le -1.61) \approx 0.05370.

\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}.

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