1. The range of guest on New Year's Eve was 15>
2 answers:
Answer:
1.False
2. True
3.) True
4. False
5.) False
6.) True
Step-by-step explanation:
1.)
Range = maximum - minimum
New year's eve :
Maximum = 22 ; minimum = 3
Range = 22 - 3 = 19
2.)
Median for Thanksgiving = 17
50% up to 17 ;hence the remaining 50% is also 17 and above
4.)
IQR for Thanksgiving
Q3 - Q1
23 - 12 = 11
5.)
The median number of people at new year's eve for dinner was 7 (line in between the box)
6.)
3/4th of new year eve's dinner = Q3 = 15
1/4 = Q1 ; have up to 5 people
Therefore ; the remaining (1 - 1/4) = 3/4 have 5 and above
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Answer:
see below
Step-by-step explanation:
If we let X represent the number of bagels produced, and Y the number of croissants, then we want ...
(a) Max Profit = 20X +30Y
(b) Subject to ...
6X +3Y ≤ 6600 . . . . . . available flour
X + Y ≤ 1400 . . . . . . . . available yeast
2X +4Y ≤ 4800 . . . . . . available sugar
_____
Production of 400 bagels and 1000 croissants will produce a maximum profit of $380.
__
In the attached graph, we have shaded the areas that are NOT part of the solution set. (X and Y less than 0 are also not part of the solution set, but are left unshaded.) This approach can sometimes make the solution space easier to understand, since it is white.
The vertex of the solution space that moves the profit function farthest from the origin is the one we are seeking. The point that does that is (X, Y) = (400, 1000).
