The dot plots below show rainfall totals in the Highlands and Lowlands areas of a certain region. 2 dot plots with number lines going from 2 to 32 in increments of 2. A dot plot titled Highlands. There are 2 dots above 2, 3 above 4, 2 above 6, 3 above 8, 4 above 10, 3 above 12, 4 above 14, 5 above 16, 5 above 18, 4 above 20, 2 above 22, 1 above 24, 2 above 26, 2 above 28, 1 above 30 and 1 above 32. A dot plot titled Lowlands. There are 3 dots above 2, 5 above 4, 3 above 6, 3 above 8, 5 above 10, 7 above 12, 6 above 14, 4 above 16, 2 above 18, 1 above 20, 2 above 22, 1 above 24, 0 above 26, 2 above 28, and 0 above 30 and 32. Which statement comparing the shapes of the dot plots is true? Both the Highlands and the Lowlands data points are evenly distributed around the center. Both the Highlands and the Lowlands data points are clustered toward the left of the plot. The Highlands data points are evenly distributed around the center, while the Lowlands data points are clustered toward the left of the plot. The Highlands data points are clustered toward the left of the plot, while the Lowlands data points are evenly distributed.

b

Step-by-step explanation: cause I know the answer/ i did the quiz

C. The Highlands data points are evenly distributed around the center, while the Lowlands data points are clustered toward the left of the plot.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: