1 Remove parentheses.
2(5c−7)≥c−3
2 Expand.
10c−14 ≥ c −3
3 Add 14 to both sides.
10c-14+14≥ c − 3+14
4 Simplify c−3+14 to c+11
10c≥c+11
5 Subtract c from both sides.
10c−c≥11
6 Simplify 10c−c to 9c
9c≥11
7 Divide both sides by 9.
c ≥11/9
Probability of finding no defective bulb
=95/100*94/99....86/91
=95!/(85!) / (100!/90!)
= 110983/190120
=>
Probability of finding at least one defective bulb
=1-110983/190120
= 79137/190120
= 0.41625 (to 5 decimal places)
The answer is 10.5 pounds. No there is no other solutions.
24 divided by 1/4 = 6 lbs
24 divided by 1/8 = 3 lbs
24 divided by 1/16 = 1.5 lbs
6+3+1.5= 10.5 lbs of food
The answer would be 782 because you would round down.
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS = = 57.5°
Now, tan(57.5°) =
⇒ 1.5697 =
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) =
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = () × (ST)
= () × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²