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Anastasy [175]
3 years ago
5

You manage an ice cream factory that makes two flavors: Creamy Vanilla and Continental Mocha. Into each quart of Creamy Vanilla

go 2 eggs and 3 cups of cream. Into each quart of Continental Mocha go 1 egg and 3 cups of cream. You have in stock 350 eggs and 600 cups of cream. You make a profit of $3 on each quart of Creamy Vanilla and $2 on each quart of Continental Mocha. How many quarts of each flavor should you make to earn the largest profit? HINT (See Example 2.] (If an answer does not exist, enter DNE.) Creamy Vanilla quarts Continental Mocha quarts
Mathematics
1 answer:
nadezda [96]3 years ago
3 0

Answer:

You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.

Step-by-step explanation:

This problem can be solved by a system of equations.

The largest profit is going to earned when all the eggs and cups of cream in stock are used.

I am going to call x the number of quarts of Creamy Vanilla and y the number of quarts of Continental Mocha.

The problem states that each quart of Creamy Vanilla uses 2 eggs and each quart of Continental Mocha uses 1 egg. There are 350 eggs in stock, so:

2x + y = 350

Each quart of Creamy Vanilla uses 3 cups of cream, as does each quart of Continental Mocha. There are 600 cups of cream in stock.

So:

3x + 3y = 600 Simplifying by 3.

x + y = 200

We have the following system

1)2x + y = 350

2)x + y = 200

I am going to multiply 2) by (-1) and then add with 1), so i can eliminate y

1)2x + y = 350

2)-x - y = -200

2x - x + y - y = 350 - 200

x = 150

x + y = 200

y = 200 - 150 = 50

You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.

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hoa [83]

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = \dfrac{1}{2}

Number of unfair dice = 1

Probability of getting a unfair dice  P(E₂) = \dfrac{1}{2}

Probability of getting a 3 for the fair dice P(A|E₁)= \dfrac{1}{6}

Probability of getting a 3 for the unfair dice P(A|E₂) = \dfrac{1}{3}

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83

Hence, our required probability is 0.83.

8 0
3 years ago
Use the zero product property to find the solutions to the equation x2 – 15x – 100 = 0.
IrinaK [193]

Answer:

x= 20      x =-5

Step-by-step explanation:

x^2 – 15x – 100 = 0.

What two numbers multiply to -100 and add to -15

-20 * 5 = -100

-20 +5 = -15

(x-20) (x+5) =0

Using the zero product property

x-20 =0     x+5 = 0

x= 20      x =-5

7 0
3 years ago
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Answer:  three sets of value show 3 consecutive increases and they could be the intensities during fourth, fifth, and six visits:

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Explanation:


1) The program recommends a constant intensity for 3 visits, which is what the table shows:

Day        Intensisty

1             63%

2            equal ⇒ 63%

3            equal ⇒ 63%


2) Hence, you have to determine the valid sets that meet the recommendation for the fourth, fifth, and six visits, which are the next three.


2) For the next three visits, the program recommensd increasing intensities.


There are three options that show 3 consecutive increases; they are:

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Therefore, those are the choices that apply.

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3 years ago
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zavuch27 [327]

Answer:

  h, j2, f, g, j1, i, k, l (ell)

Step-by-step explanation:

The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).

We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).

The horizontal asymptotes are ...

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  • g(x): x^2/x^2 = 1
  • j2(x): 3x^2/-x^2 = -3
  • f(x): 0x^2/(12x^2) = 0
  • k(x): 5x^2/x^2 = 5

So, the ordering least-to-greatest is ...

  h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)

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Answer:

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