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zaharov [31]
3 years ago
6

Write a function in any form that would match the graph shown below.

Mathematics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

  y = 2(x +1)(x -4)

Step-by-step explanation:

The intercepts are at grid points, so it is easiest to make use of those. For an x-intercept of "a", one of the factors of the function is (x -a). Here, the x-intercepts are -1 and +4, so the function has factors (x +1)(x -4).

When x=0, the product of these factors is (1)(-4) = -4, so there is a vertical scale factor of 2 to make the y-intercept be -8. The function can be written as ...

  y = 2(x +1)(x -4)

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Answer:

105 cm²

Step-by-step explanation:

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Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the data and then​ (e) answer the given question. Listed below
mafiozo [28]

Answer:

a) \bar X = 369.62

b) Median=175

c) Mode =450

With a frequency of 4

d) MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

<u>e)</u>s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

Replacing we got:

\bar X = 369.62

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Median=175

Part c

The mode is the most repeated value in the sample and for this case is:

Mode =450

With a frequency of 4

Part d

The midrange for this case is defined as:

MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

Part e

For this case we can calculate the deviation given by:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

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