Answer:
44
Step-by-step explanation:
Let x and y be the total number of questions in part 1 and part 2 exams respectively.
For part 1:
Number of correct question = 4/5 of the total number of question in part 1
For part 2:
Number of corrected question = 2/3 of total number of question in part 2
So, the total umbers of the questions on Stephanie's exam
=x+y
=20+24
=44
Hence, the total umbers of the questions on Stephanie's exam was 44.
The answer for the question above is TRUE.
Given a line segment with endpoints A and B, we must merely find a point on the perpendicular bisector of the line AB, then use the perpendicular line construction. To find the point on the perpendicular bisector, we must merely find a point that is equidistant from both A and B. We can do this with a compass by choosing a compass length greater than half the length AB, and then making an arc from A above the line AB. Then we make an arc from B having the same length.
<span>Since we used the same length for each arc, the intersection of the arcs, P, is equidistant from A and B. Now we simply use the perpendicular line construction for line AB and P, which is off the line. This line will be a perpendicular bisector. </span>
Answer:
4000
Step-by-step explanation:
36kg = 36000g
36000g/9 bags = 4000g
Answer: 4/25
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Explanation:
Let's define two events A and B
A = event of selecting red on the first draw
B = event of selecting red on the second draw
P(A) is the notation that means "probability of event A occurring"
P(A) = 2/5 because there are 2 red marbles out of 5 total (2 red + 3 black = 5)
Similarly, P(B) = 2/5 as well because A and B deal with the same color red, and because Abby put the first marble back
Multiply the probabilities
P(A and B) = P(A)*P(B) ... see note below
P(A and B) = (2/5)*(2/5)
P(A and B) = (2*2)/(5*5)
P(A and B) = 4/25 which is the answer
Note: the equation used is only valid if events A and B are independent, which they are in this case. The fact we put the marble back means the chances of picking red are the same as before.
