Answer:
49.6cm
Step-by-step explanation: First you have to find the diameter of the circle by multiplying the radius by 2.
8x2=16
Then multiply 16 by 3.1
16x3.1=49.6
Step-by-step explanation (Question 1):
<u>Step 1: (Question 1): Subtract x from both sides.</u>
5x+3=x+13
5x+3−x=x+13−x
4x+3=13
<u>Step 2: (Question 1): Subtract 3 from both sides.</u>
4x+3−3=13−3
4x=10
<u>Step 3: (Question 1): Divide both sides by 4.</u>
4x/4 = 10/4
FIRST ANSWER: x = 5/2
Step-by-step explanation (Question 2):
<u>Step 1: (Question 2): Multiply by LCM</u>
15x^2 - 6 = x
<u>Step 2: (Question 2): Solve 15x^2</u>
15x^2 - 6 = x:
x = 2/3
x = -3/5
<u>Step 3: (Question 2): Solve</u>
ANSWER: x=0.666667 or x=−0.6
See Attachment 1 for question 1 steps (FULL)
See Attachment 2 for question 2 steps (FULL)
Answer:
1) x = 5/2
2) x=0.666667 or x=−0.6
Hope this helps.
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
