Find the median data 46,22,46,49,45,39,40,42,41,39,40,36
ddd [48]
Answer:
median is 40.5
Step-by-step explanation:
22, 36, 39, 39, 40, 40, 41, 42, 45, 46, 46, 49
half way between numbers in numerical order
lands between 40 and 41
40.5
Answer:
- The graph has a minimum.
- The graph has a y-intercept at (0, -3).
- The solutions ... are -1 and 3.
- The vertex is located at (1, -4).
- In the equation, 'a' would be positive.
Step-by-step explanation:
When the graph has a low point, it has a minimum. 'a' is positive in that case. The coordinates of that low point are (1, -4). That point is the vertex.
The graph crosses the y-axis at y = -3, so the y-intercept is (0, -3).
The graph crosses the x-axis at (-1, 0) and (3, 0). These points represent the solution to the equation y = 0.
- The graph has a minimum.
- The graph has a y-intercept at (0, -3).
- The solutions ... are -1 and 3.
- The vertex is located at (1, -4).
- In the equation, 'a' would be positive.
Answer:
No each price is different
Step-by-step explanation:
4/20 gives you $5 a piece
6/37.50 gives you $6.25 per wand
10/65 gives you 6.50 per wand
These are the answer that I got:
A. <span>That means: "The more it costs, the fewer people you can invite."
B. <span>Amount you can spend = M; You will have to come up with this number - whatever you think is reasonable.
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C. (C = Cost per person, P = number of people you can invite )
D. You will have two equations. In the first equation, C will be the per-person cost at the first bowling alley. In the second equation, C will be the per-person cost at the second bowling alley.
E. <span>Be sure to think this through. Yes, you will be able to host more guests at the bowling alley with the lowest per-person cost, but is that the only factor you need to consider? What about location? Quality of lanes and bowling equipment? Music? Food? Customer Service? Consider ALL aspects.</span> You might pick the more expensive bowling alley because you feel would have a better experience, or maybe you don't want to invite lots of people anyway.<span>
Hope this helps!</span>
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