All categories

4 years ago

Please could u help me with this Thank u sooo much:)

Mathematics

1 answer:

Simora [160]4 years ago

7 0

Ok so the spread can be given by the range, interquartile range and standard deviation in this case.
Gretchen's data: Range = 22 - 3 = 19, IQR = 5, Sx = 4.69
Manuel's data: Range = 14 - 3 = 11, IQR = 4, Sx = 3.14

1. Gretchen's data is more spread out than Manuel's, which is primarily displayed by the much larger range and standard deviation for Gretchen's data (range = 19, standard deviation = 4.69) compared to Manuel's data (range = 11, standard deviation = 3.14). The interquartile range of both data sets are almost equal however, with the IQR of Gretchen's data equal to 5 and that of Manuel's data equal to 4, which may show that the greater spread of Gretchen's data is affected by the larger values in his data, namely the outlier at 22.

2. Gretchen's data set with outlier removed:
Mean = 8.71
Median = 8.5
Standard deviation = 3.31
Interquartile range = 3

3. Removing the outlier moved the centre closer, with the mean now at 8.71 and median at 8.5, compared to Manuel's data's mean of 8 and median of 8. The spread was also closer to Manuel's, with the standard deviation now equal to 3.31 (Manuel's data's standard deviation = 3.14) and the range now equal to 12 (compared to Manuel's data's range of 11). The IQR range did change from 5 to 3 after removing the outlier but there is still a difference of 1 between the IQR of Gretchen's data and that of Manuel's data. Overall however, both the centre and spread moved closer to that of Manuel's data after removing the outlier.

You might be interested in

Hannah travels 6 times as many minutes to work as Raoul does. Together, they travel for 63 minutes. How many minutes does Hannah

Anna71 [15]

The equation would be 6x+x=63. Not sure how the models play out, but good luck

7 0

3 years ago

Simplify 8(x + 3) - 2x. Answer options: A) 6 x + 3 B) 7 x C) 6 x + 24

s344n2d4d5 [400]

Answer:

Expand the bracets then simplfy and then factorise

4 0

3 years ago

Read 2 more answers

Can someone help please??????

denis23 [38]

Answer:

Please refer to the attachment

8 0

3 years ago

Emily took 300 US dollars to the bank to exchange it for Canadian dollars. The exchange rate on that day was 1.25 Canadian dolla

Paha777 [63]

Emily wanted to exchange her 300 US Dollars for Canadian Dollars.

The exchange rate for it was 1.25 Canadian Dollars for 1 US Dollar.

This means that 1 US Dollar = 1.25 Canadian Dollars.

So if we were to follow this pattern,

2 US Dollars = 2 × 1.25 = 2.5 Canadian Dollars

3 US Dollars = 3 × 1.25 = 3.75 Canadian Dollars

4 US Dollars = 4 × 1.25 = 5 Canadian Dollars

So for 300 US Dollars, we'll need to multiply 300 by 1.25.

300 US Dollars = 300 × 1.25 = 375 Canadian Dollars.

Hope it helps. :)

4 0

3 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

4 0

3 years ago