Let m represent the number of children playing soccer. Those children are separated into 4 equal teams. Write and expression for
1 answer:
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Answer:
∠EFD ≅ ∠EGD ⇒ A
Arc ED ≅ arc FD ⇒ C
m arc FD = 120° ⇒ E
Step-by-step explanation:
Let us revise some facts
- Equal chords subtended equal arcs
- The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
- The measure of a central angle is equal to the measure of its subtended arc
- If one angle of an isosceles triangle measure 60° then the triangle is equilateral
- The sum of the measures of the interior angles of any quadrilateral is 360°
In the quadrilateral CDGE
∵ m∠G = 60°
∵ m∠GDC = m∠GEC = 90°
- By using the 5th rule above
∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°
∴ 60 + 90 + m∠DCE + 90 = 360
∴ 240 + m∠DCE = 360
- Subtract 240 from both sides
∵ m∠DCE = 120°
In circle C
∵ ∠DCE is a central angle subtended by arc DE
∵ ∠DFE is an inscribed angle subtended by arc DE
- By using the 2nd rule above
∴ m∠DFE = m∠∠DCE
∵ m∠DCE = 120°
∴ m∠DFE = (120)
∴ m∠DFE = 60°
- That means ∠EFD ≅ ∠EGD because their measure is 60°
∴ ∠EFD ≅ ∠EGD
In Δ EFD
∵ EF = FD
∵ m∠DFE = 60°
- By using the 4th rule above
∴ Δ EFD is an equilateral triangle
∴ ED = FD = FE
In circle C
∵ Side ED subtended by arc ED
∵ Side FD subtended by FD
∵ Side ED ≅ side FD ⇒ proved
- By using the 1st rule above
∴ Arc ED ≅ arc FD
∵ m∠ECD = 120°
∵ ∠ECD is a central angle subtended by arc ED
- By using the 3rd rule above
∴ m∠ECD = m arc ED
∴ m of arc ED = 120°
∵ Arc ED ≅ arc FD
∴ m arc ED = m arc FD
∴ m arc FD = 120°
Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:
a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:
b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.
c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.