a new business began tracking how many customers entered the store each week. the store owner realized the…

a new business began tracking how many customers entered the store each week. the store owner realized the relationship between the week and the number of customers was quadratic. if the trend continues, during which week will 156 customers enter the store?\na week 7\nb week 8\nc week 16\nd week 17

a new business began tracking how many customers entered the store each week. the store owner realized the relationship between the week and the number of customers was quadratic. if the trend continues, during which week will 156 customers enter the store?\na week 7\nb week 8\nc week 16\nd week 17

Answer

  1. Assume the quadratic - function model:
    • Let the quadratic function be (y = ax^{2}+bx + c), where (y) represents the number of customers and (x) represents the week number. Since we don't have the specific function, we can also use the fact that for a quadratic equation (y = ax^{2}+bx + c), we can solve for (x) when (y) is given. In this case, (y = 156).
    • However, if we assume a simple quadratic - like situation where we might have a table of values or some basic pattern. Let's assume we can try to substitute the values of (x) (week number) into a quadratic - like relationship.
    • Let's assume we have no other information about the quadratic function, and we just test the values one by one.
  2. Test option A:
    • If (x = 7), assume a general quadratic function (y=ax^{2}+bx + c). Without loss of generality, if we assume (a = 1), (b = 0), (c = 0) (a very simple quadratic (y=x^{2})), then (y=7^{2}=49). This is much less than 156.
  3. Test option B:
    • If (x = 8), then (y = 8^{2}=64). This is also less than 156.
  4. Test option C:
    • If (x = 16), then (y = 16^{2}=256). This is greater than 156.
  5. Test option D:
    • If (x = 17), then (y = 17^{2}=289). This is also greater than 156.
    • Let's assume a more general quadratic function (y = ax^{2}+bx + c). We know that the quadratic function is a parabola. If we assume the vertex of the parabola is not in the range of our interest and the function is increasing (since the number of customers is growing in a quadratic - like way).
    • We can also try to solve the quadratic equation (ax^{2}+bx + c=156). But since we have no information about (a), (b), and (c), we can use the fact that we can estimate.
    • Let's assume a simple quadratic (y = x^{2}). We want to find (x) such that (x^{2}=156), then (x=\sqrt{156}\approx12.49). Since the actual quadratic function may have non - zero (b) and (c) values, we need to test the closest values.
    • If we assume a quadratic function that is increasing with (x), we note that as (x) increases, (y) increases.
    • Let's assume we have some data - fitting quadratic function. If we test the values one by one, we find that if we assume a quadratic growth pattern, when we calculate for each option:
      • For (x = 7), (y) (in a simple quadratic (y = x^{2})) is (49).
      • For (x = 8), (y) is (64).
      • For (x = 16), (y) is (256).
      • For (x = 17), (y) is (289).
      • We can assume that the quadratic function is increasing. We know that the value of (x) (week number) should be between 12 and 13 (since (\sqrt{156}\approx12.49)). Among the given options, the closest value to the solution of the quadratic equation (y = 156) (for the quadratic relationship between week and number of customers) is when we consider the growth pattern. If we assume the quadratic function is increasing, we note that as (x) increases, (y) increases. Since (12.49) is closer to 13, and among the options, the next available value after a value close to 13 is 16.

Answer:

C. week 16