4 cupcakes because if you do 12/ 1/.. you do kcf = to 12 x 1/3= 4
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A) The variable on the horizontal axis of the graph (the independent variable) is "pounds of rice". That is what the first number in the ordered pair (6, 18) represents.
The variable on the vertical axis of the graph (the dependent variable) is "total cost in dollars". That is what the second number in the ordered pair represents.
(6, 18) represents that the total cost of purchasing 6 lbs of rice is $18.
B) The unit price is found at the point where the independent variable has the value 1. That would be at the point (1, 3), which indicates the unit price is $3 per pound.
C) You would have to buy 4 lbs of rice for the total cost to be $12. There are at least two ways to find the answer.
- Draw a horizontal line on the graph at cost = $12. It intersects the graph at lbs = 4.
- Divide the total cost by the unit price. $12/($3/lb) = 4 lb.
Would it be 9 because 8 times 3 equals 24 and 3 times 3 equals 9
The third choice is appropriate.
an = 5 - 3(n - 1); all integers n ≥ 1
_____
This equation follows the form for the general term of an arithmetic sequence.
an = a1 + d(n - 1)
where a1 is the first term (corresponding to n=1), and d is the common difference. From the problem statement, a1 = 5 and d = 2 - 5 = -3.
Answer:
1727 students
Step-by-step explanation:
Here we have the formula for sample size given as

Where:
p = Mean
ME = Margin of error = 3
z = z score
Therefore, we have
p = 150/240 = 0.625
z at 99 % = 2.575
ME =
3%
Therefore 
The number of students Professor York have to sample to estimate the proportion of all Oxnard University students who watch more than 10 hours of television each week within ±3 percent with 99 percent confidence = 1727 students.