Answer:
a) E(X) = 3.30
b) If these are the only two grades (A and B) available in the school, then this calculated expected value will truly be the expected GPA for each quarter or semester But, if like real life, there are other grades available at the school not included in this calculation, then this value seizes to be the expected value for the GPA over any time period.
Step-by-step explanation:
P(X=xᵢ) = pᵢ
For an A,
P(X=4.0) = 0.3
For a B
P(X=3.0) = 0.7
But E(X) = Σ xᵢpᵢ
E(X) = (0.3 × 4.0) + (0.7 × 3.0) = 3.30
b) If these are the only two grades available in the school, then this calculated expected value will truly be the expected GPA for each quarter or semester But, if like real life, there are other grades available at the school not included in this calculation, then this value seizes to be the expected value for the GPA over any time period.
Answer:
-2 C < 4 C
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
split it in half u get a subtract sing makes that (8-4=4)
a
I searched and found this figure online.
I'm guessing there's additional information that OM perpendicular to CD and EF perpendicular to ON.
The diameters meeting chords at right angles also bisect the chords, so
CM = (1/2) CD = 28
The radius OC makes the hypotenuse of a right triangle with legs OM and CM
r² = OM² + CM² = 20² + 28² = 1184
r = 4√74
a Answer: 4√74
b
FN is the leg of a right triangle with other leg ON=16 and hypotenuse OF=r because it's a radius.
FN² = r² - ON² = 1184 - 16² = 1184 - 256 = 928
FN = √928
b Answer: √928
c
EF is twice FN by the perpendicular diameter bisecting a chord theorem.
EF = 2√928
Tenths?
c Answer: 60.9