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dybincka [34]
3 years ago
9

Following a state-wide boom in voter registration and participation, Lakeview's election board claimed that the increased partic

ipation in its precinct resulted from the new busing campaign that brought many voters to their neighborhood polling station on Election Day.
Which of the following, if it could be carried out, would be most useful in an evaluation of Lakeview's assessment of the causes of the increased voter participation in its precinct?
A) Determining the average increase in voter participation in each neighborhood of Lakeview.B) Using polling data to predict the participation rates for the upcoming election in Lakeview's precinct.C) Comparing the total number of voters participating when the busing campaign brought them to their neighborhood polling station to the total number of voters participating before the busing campaign.D) Comparing the increase in voter participation in Lakeview to the increase in participation in precincts demographically similar to Lakeview that did not implement busing campaigns.E) Comparing the long-term impact of the busing campaign on Lakeview's voter registration patterns with the short-term impact of the busing campaign on Lakeview's voter participation.
Social Studies
1 answer:
Natasha_Volkova [10]3 years ago
6 0

Answer:

.D) Comparing the increase in voter participation in Lakeview to the increase in participation in precincts demographically similar to Lakeview that did not implement busing campaigns.

Explanation:

The is called a posttest-only control group design

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Q1. sin27°   = x/8

Solution:

We have to solve for x, therefore, we will rearrange the given equation for x.

We get,
x = 8 × sin27°

Using the calculator,

sin27° = 0.45

Now substitute the value of sin27° into the main equation.

we get,
x = 8 × 0.45
x = 3.63 (rounded to the nearest hundredth)


Q2. tan 18°  = n / 75

Solution:
We have to solve for n, therefore, we will rearrange the given equation for n.
We get,
n = 75 × tan 18°
Using the calculator,
tan 18° = 0.32
Now substitute the value of tan 18° into the main equation.
we get,
x = 75 × 0.32
x = 24.37 (rounded to the nearest hundredth)

Q3. sin40°  = 4 / a

Solution: We have to solve for a, therefore, we will rearrange the given equation for a.
We get,
a = 4 ÷ sin40°
Using the calculator,
sin40° = 0.64
Now substitute the value of sin40° into the main equation.
we get,
a = 4 ÷ 0.64
a = 6.25 (rounded to the nearest hundredth)

Q4. cos5°   = 92 / y

Solution: We have to solve for y, therefore, we will rearrange the given equation for y.
We get,
y = 92 ÷ cos5°
Using the calculator,
Cos5° = 0.99
Now substitute the value of cos5° into the main equation.
we get,
y = 92 ÷ 0.99
y = 92.92 (rounded to the nearest hundredth)

Q5:
Given the shape attached, therefore, using the triangle given, we have:
Angle of elevation = 35°
length of Opposite side to the angle = x
Length of Hypoteneus = 12
Calculations:
Using the SOH CAH TOA rules:
SOH stands for SineФ = Opposite ÷ Hypotenuse.

CAH stands for CosineФ = Adjacent ÷ Hypotenuse.

TOA stands for TangentФ = Opposite ÷ Adjacent.

Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine35° = x ÷ 12

               0.5735  = x ÷ 12

                          x = 0.5735 × 12

                          x = 6.88 (rounded to the nearest hundredth)

Q6: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 54°
length of the adjacent side to the angle = x
Length of Hypoteneus = 30
Calculations:
Using the SOH CAH TOA rules:

Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos54° = x ÷ 30

               0.5877  = x ÷ 30

                          x = 0.5877 × 30

                          x = 17.63 (rounded to the nearest hundredth)

Q7: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 22°
length of the adjacent side to the angle = 85
length of the opposite side to the angle = x
Calculations:

Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan22° = x ÷ 85

              0.4040 = x ÷ 85

                          x = 0.4040 × 85

                          x = 34.34 (rounded to the nearest hundredth)

Q8: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 16°
length of the opposite side to the angle = x
Length of Hypoteneus = 14
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine16° = x ÷ 14

               0.2756  = x ÷ 14

                          x = 0.2756 × 14

                          x = 3.86 (rounded to the nearest hundredth)

Q9: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 65°
length of the adjacent side to the angle = 9
length of the opposite side to the angle = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = x ÷ 9

               2.1445 = x ÷ 9

                          x = 2.1445 × 9

                          x = 19.30 (rounded to the nearest hundredth)

Q10: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 51°
length of the adjacent side to the angle = x
Length of Hypoteneus = 70
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos51° = x ÷ 70

              0.6293  = x ÷ 70

                          x = 0.6293 × 70

                          x = 44.05 (rounded to the nearest hundredth)

Q11: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 36°
length of the opposite side to the angle = 15
Length of Hypoteneus = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine36° = 15 ÷ x

               0.5877  = 15 ÷ x

                          x = 15 ÷ 0.5877

                          x = 25.52 (rounded to the nearest hundredth)

Q12: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 60°
length of the adjacent side to the angle = x
length of the opposite side to the angle = 100

Calculations:
Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = 100 ÷ x

               2.1445 = 100 ÷ x

                          x = 100 ÷ 2.1445

                          x = 46.63 (rounded to the nearest hundredth)

Q13: When a 25-ft ladder is leaned against a wall, it makes a 72° with the ground. How high up on wall does the ladder reach?

Solution: Given the shape attached, therefore, using the triangle given, we have:

The angle of elevation from the ground = 72°
length of the wall opposite to the angle = X
Length of ladder (Hypoteneus) = 25 feet

Calculations:
Using the SOH CAH TOA rules:
Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine72° = x ÷ 25

               0.9510  = x ÷ 25

                          x = 25 ÷ 0.9510

                          x = 23.77 (rounded to the nearest hundredth)


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