Select the correct answer from each drop-down menu. In the figure, and ∠EIA ≅ ∠GJB. Complete the following statements to prove that ∠IKL ≅ ∠DLH. ∠EIA ≅ ∠IKC and ∠GJB ≅ ∠JLD because they are corresponding angles of parallel lines cut by a transversal. So, if ∠EIA ≅ ∠GJB, then ∠IKC ≅ ∠JLD by the . ∠IKL and ∠IKC and ∠DLH and ∠JLD are pairs of supplementary angles by the . m∠IKL + m∠IKC = 180° (1) ∠IKC ≅ ∠JLD, so m∠IKC = m∠JLD (2) Applying the to equations (1) and (2), we get m∠IKL + m∠JLD = 180°. Therefore, ∠IKL and ∠JLD are supplementary angles. We've already shown that ∠DLH and ∠JLD are supplementary angles. Therefore, ∠IKL ≅ ∠DLH by the .

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.