The price of a bag of candy is a linear function of the amount of candy in the bag. A bag containing 25 pieces costs $0.99 and a
1 answer:
Answer:
About 152 pieces of candy
Step-by-step explanation:
Set up equal proportions to solve for the $5.99 bag.
We know that 24 pieces = .99, so
Make sure to keep the pieces on top, and the dollars on bottom (or vice versa). If your units are messed up, it will not work.
Cross-multiply.
25(5.99) = .99x
Simplify.
149.75 = .99x
Divide both sides by .99.
x = 151.26 = about 151 pieces of candy
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The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
Answer:
A 90° counterclockwise rotation about the origin and then a translation 16 units right and 16 units up
Solution -Rotating the triangle 90° counterclockwise will take the triangle to 3rd quadrant and then further moving it 16 steps right will take it to 4th quadrant and followed by 16 steps upward will take it to the desired position which is in 1st quadrant.
