Answer:
750 < p < 1500
Step-by-step explanation:
The total cost of the affair for p attendees is ...
750 +2.25p
The average cost is that number divided by p. The answer choices suggest that "between" means that the cost per person cannot be 2.75 or 3.25. Applying the limits to the average cost, we get ...
2.75 < (750 +2.25p)/p < 3.25
2.75 < 750/p + 2.25 < 3.25 . . . . . . . . separating the fraction parts
0.50 < 750/p < 1.00 . . . . . . . . . . . . . . subtract 2.25
2 > p/750 > 1 . . . . . . . . . . . . . . . . . . . . . take the reciprocal*
1500 > p > 750 . . . . . . . . matches choice 4
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* Taking the reciprocals of numbers with the same sign reverses their order:
2 < 3 but 1/2 > 1/3
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An alternate approach to the solution would be to multiply by p:
0.50p < 750 < 1.00p
To solve this requires it be made into two inequalities:
0.50p < 750 ⇒ p < 1500
and
750 < 1.00p ⇒ 750 < p
Then these would need to be recombined to form the answer:
750 < p < 1500
67 is the correct answer trust me i just answered and got it correct.
Answer: Choice C
Graph with filled in circles at d = 0 and d = 2, shading in between
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Explanation:
d = amount downloaded in gigabytes
The smallest amount is d = 0. We cannot download a negative amount of data, so this is why d = 0 is the smallest.
The largest amount allowed is d = 2. This is the cap that the ISP has set up.
So basically d can be anything between d = 0 and d = 2, including both endpoints. This means
We use filled in circles for both endpoints to show to the reader "include these endpoints". Shading is done in between to show the entire solution set of possible d values. For instance d = 1 is in that region so it is possible to have this solution. Something like d = 4 is outside the region and not possible.
61/100, because if it is 0.(value) than it is value / 100 assuming value is 2 digits long. We cannot simplify it further because 61 is prime number.
The distance between B and B' will be the same as distance between A and A'.
= sqrt ( (5-1)^2 + ( 1 - -2)^2)
= sqrt (16 + 9)
= 5 units answer
