Answer: C. Significant at 0.036
Step-by-step explanation:
Given:
Total samples selected Ns= 500
Airplanes that arrived on time Na = 482.
Airplanes that arrived late Nl = 500 - 482 = 18
Calculating the probability of an airplane arriving late:
P(L) = Nl/Ns
P(L) = 18/500
P(L) = 0.036
An event is deemed significant if its probability is equal to or less than 0.05.
As P(L) < 0.05
P(L) = Significant at 0.036
Answer:
Angle x = angle 115°.
Step-by-step explanation:
Given: We have two parallel lines intersected by a transversal.
To find: The value of x.
Solution: Since two parallel lines are intersected by a transversal, corresponding angles, which are a pair lying on the same side of the transversal—one being on the interior and the other on the exterior—are equal.
Thus, angle x = angle 115, being corresponding angles.
Therefore, angle x = angle 115.
The outcome is "
GK=37, GH=14, and JK=19".The problem references a missing attachment file, thus please find it attached.
The provided values are:
GJ=4x+2
HK=6x-1
Given GK =37
Need to calculate:
HJ=?
From GK: GK = GJ+JK yields GJ+(HK-HJ)
= 4x+2+((6x-1)-x) yields
9x+1...(a)Substituting (a) into the equation gives: 9(4)+1 yields 37 solves to x=4... for answer (b) and (c). Substituting x=4 yields:
GH=14 and JK=19.
Answer:
Review the explanation
Step-by-step explanation:
The triangles ΔABC and ΔBAD are congruent, establishing:
- AB ≅ BA;
- AC ≅ BD;
- BC ≅ AD;
- ∠ABC ≅ ∠BAD;
- ∠BCA ≅ ∠ADB;
- ∠CAB ≅ ∠DBA.
Now, consider triangles AEC and BED. In these triangles, we have:
- AC ≅ BD;
- ∠EAC ≅ ∠EBD (due to ∠CBA ≅ ∠BAD);
- ∠AEC ≅ ∠BED (being vertical angles).
Therefore, ΔAEC ≅ ΔBED, which leads to the conclusion that:
AE ≅ EB.
This indicates that line segment CD bisects segment AD.
