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kvv77 [185]
3 years ago
7

The entry fee to an Amusement Park is $125 and the cost of each ride is $ 7.25 ( Mon - Thurs) and $ 8.50 ( Fri - Sun). Ava's mom

gave her $ 250 to go to the park on Friday. Which inequality represents the given situation?
a. 7.25r ≤ 205


b. 8.50r ≤ 20


c. 7.25r + 125≤250


d. 8.50r+ 125 ≤ 250
Mathematics
1 answer:
icang [17]3 years ago
8 0

Answer:

D

Step-by-step explanation:

Well it is Friday so its 8.5 each ride.

8.5r which is basically how many rides she can go on after spending 125 on the admission

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part A: Mean:6.42 Median:7

Part b: median

Part c: no it’s skewed left and unimodal

Step-by-step explanation:

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The table below shows two equations:
CaHeK987 [17]

Answer:

The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution.

Step-by-step explanation:

Rearranging the two equations, you get ...

  • |4x -3| = 9 . . . . . has two solutions
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The above-listed answer is the only one that matches these solution counts.

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Testing the above values of x reveals they are, indeed, solutions to Equation 1.

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The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

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Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

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However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

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However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

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