ANSWER:
Change in the Independent Variable / Change in the Dependent Variable
hope this helps!!
Problem 44
The term "bisect" means "cut in half".
Since BD bisects angle ABC, this means the smaller angles ABD and DBC are congruent.
angle ABD = angle DBC
x+15 = 4x-45
15+45 = 4x-x
60 = 3x
3x = 60
x = 60/3
x = 20
<h3>Answer: x = 20</h3>
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Problem 45
We use the same idea as the previous problem
angle ABD = angle DBC
2x+35 = 5x-22
35+22 = 5x-2x
57 = 3x
3x = 57
x = 57/3
x = 19
<h3>Answer: x = 19</h3>
Answer:
1 ft : 2.5 yd or 1 ft : 2 1/2 yd
Step-by-step explanation:
He wants a real dimension of 10 yards to be represented by a scale dimension of 4 ft.
4 ft : 10 yd
Divide both sides by 4.
1 ft : 2.5 yd or 1 ft : 2 1/2 yd
Answer:
(E) 0.71
Step-by-step explanation:
Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.
So, the probability that the student has a GPA lower than 3.5 and is not taking any AP classes is calculated as:
P(A'∩B') = 1 - P(A∪B)
it means that the students that have a GPA lower than 3.5 and are not taking any AP classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.
Therefore, P(A∪B) is equal to:
P(A∪B) = P(A) + P(B) - P(A∩B)
Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled in at least one AP class is 0.12
So, P(A∪B) is equal to:
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = 0.25 + 0.16 - 0.12
P(A∪B) = 0.29
Finally, P(A'∩B') is equal to:
P(A'∩B') = 1 - P(A∪B)
P(A'∩B') = 1 - 0.29
P(A'∩B') = 0.71
