Answer:
Step-by-step explanation:
It's given in this question,
m∠2 = 41°, m∠5 = 94° and m∠10 = 109°
Since, ∠2 ≅ ∠9 [Alternate interior angles]
m∠2 = m∠9 = 41°
m∠8 + m∠9 + m∠10 = 180° [Sum of angles at a point of a line]
m∠8 + 41 + 109 = 180
m∠8 = 180 - 150
m∠8 = 30°
Since, m∠2 + m∠7 + m∠8 = 180° [Sum of interior angles of a triangle]
41 + m∠7 + 30 = 180
m∠7 = 180 - 71
m∠7 = 109°
m∠6 + m∠7 = 180° [linear pair of angles]
m∠6 + 109 = 180
m∠6 = 180 - 109
= 71°
Since m∠5 + m∠4 = 180° [linear pair of angles]
m∠4 + 94 = 180
m∠4 = 180 - 94
m∠4 = 86°
Since, m∠4 + m∠3 + m∠9 = 180° [Sum of interior angles of a triangle]
86 + m∠3 + 41 = 180
m∠3 = 180 - 127
m∠3 = 53°
m∠1 + m∠2 + m∠3 = 180° [Angles on a point of a line]
m∠1 + 41 + 53 = 180
m∠1 = 180 - 94
m∠1 = 86°
Answer:
I am not able to login to the account can u please type the question
<span>The computation for the confidence level = (18/20) x 100% =
90 %, the E or the margin error = 0.023 and the p = to 0.71. Confidence
interval can be achieved by using the formula (p-E, p+E) = (0.71-0.025,
0.71+0.025). Therefore, the confidence interval is (0.685,0.735).</span>
Answer:
8 nickels and 9 dimes
Step-by-step explanation:
If all were dimes, the value would be $1.70. It is $0.40 less than that. Changing a dime for a nickel reduces the value by $0.05, so there must have been $0.40/$0.05 = 8 such changes.
There are 8 nickels and 9 dimes.
_____
<em>Check</em>
8 · 0.05 + 9 · 0.10 = 0.40 + 0.90 = 1.30 . . . the answer checks OK
I think around 44% chance that all three flights would arrive on time, but i'm not that sure.
