We know that
If a tangent segment and a secant segment are drawn to a <span>circle </span><span>from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment
</span>so
AB²=BD*BC
BD=12
BC=12+15----> BC=27
AB²=12*27----> 324
AB=√324----> AB=18
the answer is
AB=18 units
Answer:
AC = √105 ≈ 10 cm (nearest whole number)
Step-by-step explanation:
Use Pythagoras' Theorem: a² + b² = c²
(where a and b are the legs, and c is the hypotenuse of a right triangle)
Give one of the legs = 4cm and the hypotenuse = 11 cm:
⇒ 4² + AC² = 11²
⇒ 16 + AC² = 121
⇒ AC² = 121 - 16
⇒ AC² = 105
⇒ AC = ±√105
As distance is positive, length of AC = √105 ≈ 10 cm (nearest whole number)
Given Information:
Probability of super event = P(S) = 0.0023
Number of suppliers = n = 3
Probability of unique event = P(U) = 0.014
Required Information:
Probability that all three suppliers will be disrupted = ?
Answer:
P(3) = 0.0023
Step-by-step explanation:
We want to find out the approximate probability that all three suppliers will be disrupted at the same time at some point during the next five years.
The required probability is given by
P(n) = P(S) + (1 - P(S))*P(U)ⁿ
Where P(S) is the probability of super event that will disrupt all suppliers, P(U) is the probability of unique event that would disrupt one of the suppliers and n is the number of suppliers.
P(3) = 0.0023 + (1 - 0.0023)*(0.014)³
P(3) = 0.0023 + (0.9977)*(0.014)³
P(3) = 0.0023
The correct option is C = 0.0023
Therefore, there is 0.23% probability that all three suppliers will be disrupted at the same time at some point during the next five years.
L=2w+2
The length is 2 times the width plus two
Answer:
18 ft
Step-by-step explanation: