Answer:
m∠ABE = 62°
Step-by-step explanation:
Since, the given quadrilateral is a kite, diagonals will intersect at 90°.
Therefore, m∠AEB = m∠CEB = 90°
m∠BAE = 28° [Given]
m∠BCE = 58° [Given]
From ΔABE,
m∠BAE + m∠BEA + m∠ABE = 180°
28° + 90° + m∠ABE = 180°
m∠ABE = 180° - 118°
= 62°
Therefore, measure of angle ABE = 62°.
Answer:
0.006
Step-by-step explanation:
Answer:
Following are the solution to the given equation:
Step-by-step explanation:
Please find the complete question in the attachment file.
In point a:




In point b:


In point c:


In point d:



Answer:
0.11069
Step-by-step explanation:
We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.
P(Y ≤ 5) = P(Y = 0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) =

I hope that works for you!
Area = 2pi*r*h + 2pi*r^2
Area = 2pi*r(h + r)
radius of 7 in and a height of 32 in
Area = 14pi*39 = 546pi in^2