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alina1380 [7]
1 year ago
14

Fancy goldfish x cost $3 each and common goldfish y cost $1 each.Graph the function y=20-3x to determine how many of each type o

f goldfish Tasha can buy for $20.
Mathematics
1 answer:
aniked [119]1 year ago
4 0

The pairs of goldfish Tasha can buy for $20 are (0,20), (1,17), (2,14),

(3,11),(4,8),(5,5),and (6,2)

Step-by-step explanation:

To find the how many type of goldfish that Tasha can buy for $20 is the set of positive ordered pairs.

Let x denote the number of fancy goldfish.

Let y denote the number of common goldfish.

To find how much goldfish can Tasha buy for $20 is to substitute the value for x in the equation y=20-3x

Now, substituting x=0 in y=20-3x, we get,

y=20

The ordered pair is (0,20)

Similarly, substituting x=1,

y=20-3(1)\\y=20-3\\y=17

For x=2,

y=20-6\\y=14

For x=3,

y=20-9\\y=11

For x=4,

y=20-12\\y=8

For x=5,

y=20-15\\y=15

For x=6,

y=20-18\\y=2

Thus, the pairs of goldfish Tasha can buy for $20 are (0,20), (1,17), (2,14),(3,11),(4,8),(5,5),and (6,2)

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6 ice cream cones cost $10. How much does 9 cost
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Points A, B, and C are on line AC. A horizontal line has points A, B, C. A line extends from point B up and to the left to point
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Davison Electronics manufactures two LCD television monitors, identified as model A andmodel B. Each model has its lowest possib
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The question is incomplete. The complete question is :

Davison Electronics manufactures two LCD television monitors, identified as model A and model B. Each model has its lowest possible production cost when produced on Davison’s new production line. However, the new production line does not have the capacity to handle the total production of both models. As a result, at least some of the production must be routed to a higher-cost, old production line. The following table shows the minimum production requirements for next month, the production line capacities in units per month, and the production cost per unit for each production line:

                      Production per unit                         Minimum production

Model        New Line        Old Line                              requirements

A                   $30                $50                                          50,000

B                   $25                $40                                           70,000

Production  80000          60000

line capacity.

Let

AN- Units of model A produced on the new production line

AO Units of model A produced on the old production line

BN Units of model B produced on the new production line

BO Units of model B produced on the old production line

Davison's objective is to determine the minimum cost production plan. The computer solution is shown in Figure 3.21.

a. Formulate the linear programming model for this problem using the following four constraints:

Constraint 1: Minimum production for model A

Constraint 2: Minimum production for model B

Constraint 3: Capacity of the new production line

Constraint 4: Capacity of the old production line  

Solution :

<u>Linear programming model</u>:

Linear programming is defined as a mathematical model where the linear function is either minimize or maximize when they are subjected to some constraints.

The linear programming model is determined as follows :

Minimum : 30 AN + 50 AO + 25 BN + 40 BO

This is subject to the constraints as :

AN+AO \geq 60,000

BN+BO \geq 70,000

AN+BN \leq 80,000

AO+BO \leq 60,000

Learn more :

https://brainly.in/question/15044395

3 0
1 year ago
If the test scores of a class of 35 students have a mean of 74.3 and the test scores of another class of 28 students have a mean
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Let the total sum of the scores of the first class of 35 students be a. The mean is 74.3 .

So 
\frac{a}{35}=74.3\\\\a=35\cdot74.3= 2600.5

Also, let the total sum of the scores of the second class of 28 students be b. The mean is 67.6 .

so 
\frac{b}{28}=67.6\\\\ b=28\cdot67.6= 1892.8



The combined group has 35+28=63 students. The sum of their scores is 

a+b=
2600.5+1892.8=4493.3


Thus, the mean of the combined group is \frac{4493.3}{63}= 71.32

6 0
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