Answer:
It will be 60 pixels high and 75 pixels wide
Step-by-step explanation:

As
- The space allowed for the mascot on a school's Web page is 75 pixels wide by 60 pixels high
- Its digital image is 500 pixels wide by 400 pixels high
So, the expression becomes








Therefore, it will be 60 pixels high and 75 pixels wide
Keywords: pixel, ratio
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Answer:
- Constraints: x + y ≤ 250; 250x +400y ≤ 70000; x ≥ 0; y ≥ 0
- Objective formula: p = 45x +50y
- 200 YuuMi and 50 ZBox should be stocked
- maximum profit is $11,500
Step-by-step explanation:
Let x and y represent the numbers of YuuMi and ZBox consoles, respectively. The inventory cost must be at most 70,000, so that constraint is ...
250x +400y ≤ 70000
The number sold will be at most 250 units, so that constraint is ...
x + y ≤ 250
Additionally, we require x ≥ 0 and y ≥ 0.
__
A profit of 295-250 = 45 is made on each YuuMi, and a profit of 450-400 = 50 is made on each ZBox. So, if we want to maximize profit, our objective function is ...
profit = 45x +50y
__
A graph is shown in the attachment. The vertex of the feasible region that maximizes profit is (x, y) = (200, 50).
200 YuuMi and 50 ZBox consoles should be stocked to maximize profit. The maximum monthly profit is $11,500.
Answer:
120 tries
Step-by-step explanation:
You can try 5 different wires in the first connection.
Then you have 4 wires left, so you try 4 wires in the second connection. Remember, you are now trying 4 wires for each of the first 5, so up to here you already made 4 * 5 tries = 20 tries.
Next you try 3, then 2, then you have 1 left.
Number of tries: 5 * 4 * 3 * 2 * 1 = 120
Answer: 120 tries
Which ordered pairs are in the solution set of the system of linear inequalities? (4, 2)
Answer: 60 of the patients are boys.
Step-by-step explanation: According to what is described, 5 in 8 patients are girls, then: 8 - 5 = 3, i.e., 3 in 8 are boys.
Mathematically the proportionality is 
If 160 patients have appointments per month and
are boys:
160.
= 60
Per month, 60 boys have appointments at the medical office.