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DedPeter [7]
3 years ago
5

1. A five-number summary of a univariate data set is determined to be [10, 15, 25, 45, 85]. These data are to be used to constru

ct a (modified) boxplot. Which of the following statements are true?
I. The mean is probably greater than the median.
II. There is at least one outlier.
III. The data are skewed to the right.

(A) I only
(B) II only
(C) III only

Mathematics
1 answer:
kompoz [17]3 years ago
3 0

Answer:

A) I only

Step-by-step explanation:

median = 25

mean = 36

Plot for given distribution is shown in fig attached below. mean is shown with red block and median with green block. plot is skewed to the left and there is no outlier.

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Line AB, segment DF, and segment CE are drawn. Segments DF and CE intersect at point Z. Which statement about the diagram is tru
Tju [1.3M]

Step-by-step explanation:

A chord is a segment with endpoints on the circle.

A tangent line is a line that touches a circle at only one point.

A diameter is a chord that passes through the center of the circle.

A secant line is a line that touches a circle at two points.

Therefore, AB is a secant line.

8 0
3 years ago
What is the midpoint of the line segment with endpoints of (-2,-2) and (4,6)?
slamgirl [31]

Answer:

option (d) is correct.

The mid points of the line segment whose ends points are (-2,-2) and (4,6) is (1,2)

Step-by-step explanation:

Given: end points of a line segment as (-2,-2) and (4,6)

We have to find the mid points of the line segment whose ends points are given.

Mid point formula is stated as ,

For a line having end points as \left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right) , the mid point can be calculated as,

\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

Here,

\left(x_1,\:y_1\right)=\left(-2,\:-2\right),\:\left(x_2,\:y_2\right)=\left(4,\:6\right)

Substitute in mid point formula, we get,

=\left(\frac{4-2}{2},\:\frac{6-2}{2}\right)

Solving further , we get,

=\left(1,\:2\right)

Thus, the mid points of the line segment whose ends points are (-2,-2) and (4,6) is (1,2)

Thus, option (d) is correct.

6 0
3 years ago
Read 2 more answers
How do you solve this?
DiKsa [7]

Triangles ABC and LBM are similar. We know this because AL and LB have the same length, so that AB is twice as long as either AL or LB. The same goes for MC and BM, and BC. The angle B is the same for both tirangles ABC and LBM, so the side-angle-side postulate tells us the triangles are similar, and in particular that triangle ABC is twice as large as LBM.

All this to say that LM must be half as long as AC, so LM has length (B) 14 cm.

3 0
3 years ago
Sylvia ran 3km 290m in the morning then she ran some more in the evening if she ran a total of 10km how far did Sylvia run in th
MAXImum [283]

Answer:

She ran 6.71km in the evening

Step-by-step explanation:

3km 290m = 3.29km

10 - 3.29 = 6.71km

4 0
3 years ago
Find the following: F(x, y, z) = e^(xy) sin z j + y tan^−1(x/z)k Exercise Find the curl and the divergence of the vector field.
natulia [17]

\vec F(x,y,z)=e^{xy}\sin z\,\vec\jmath+y\tan^{-1}\dfrac xz\,\vec k

Divergence is easier to compute:

\mathrm{div}\vec F=\dfrac{\partial(e^{xy}\sin z)}{\partial y}+\dfrac{\partial\left(y\tan^{-1}\frac xz\right)}{\partial z}

\mathrm{div}\vec F=xe^{xy}\sin z-\dfrac{xy}{x^2+z^2}

Curl is a bit more tedious. Denote by D_t the differential operator, namely the derivative with respect to the variable t. Then

\mathrm{curl}\vec F=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\D_x&D_y&D_z\\0&e^{xy}\sin z&y\tan^{-1}\frac xz\end{vmatrix}

\mathrm{curl}\vec F=\left(D_y\left[y\tan^{-1}\dfrac xz\right]-D_z\left[e^{xy}\sin z\right]\right)\,\vec\imath-D_x\left[y\tan^{-1}\dfrac xz\right]\,\vec\jmath+D_x\left[e^{xy}\sin z}\right]\,\vec k

\mathrm{curl}\vec F=\left(\tan^{-1}\dfrac xz-e^{xy}\cos z\right)\,\vec\imath-\dfrac{yz}{x^2+z^2}\,\vec\jmath+ye^{xy}\sin z\,\vec k

5 0
3 years ago
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