Answer:
0.56
it's basically the number closest to 50. Don't get tricked by the 0.05 because that's only 5% not 50%
Answer:
Step-by-step explanation:
As per Janayda,
From the figure attached,
In ΔTRQ,
m∠TRQ + m∠RQT + m∠QTR = 180°
25° + m∠RQT + 35° = 180°
m∠RQT = 180° - 60°
m∠RQT = 120°
Since, m∠RQT + m∠PQT = 180° [Linear pair of angles]
m∠PQT = 180° - m∠RQT
= 180° - 120°
= 60°
In right angled triangle TPQ,
m∠TPQ + m∠PQT + m∠PTQ = 180°
90° + 60° + m∠PTQ = 180°
m∠PTQ = 180° - 150°
= 30°
Similarly, other angles can also be evaluated from the given information.
In ΔQTP and ΔNTP,
TP ≅ TP [Reflexive property]
NP ≅ PQ [Given]
ΔQTP ≅ ΔNTP [By LL postulate for congruence]
Therefore, Janayda is correct.
While Sirr is incorrect.
Since, there is not the enough information to prove ΔRTQ and ΔMTN equal, Isabelle is incorrect.
A₅ = 1/16 and r= 1/4
Let see how to build up this formula that is going to give that term of rank n
1st term =a₁ = To be calculated
1st a₁ = a₁ x r°
2nd a₂ = a₁ x r¹
3rd a₃ = a₁ x r²
4th a₄ = a₁ x r³
5th a₅ = a₁ x r⁴
.......................
.......................
nth : a(n) = a₁ x r⁽ⁿ-¹)
Note when that the subscript of a is the same as the exponent mines 1
We know the ratio r =1/4 & the fifth term, a₅ =1/16 (given). Now let's apply the formula to calculate the unknown a₁.
a(n) = a₁ x r⁽ⁿ-¹) ==>a₅ = a₁ x (1/4)⁽⁵⁺¹⁾ ===> 1/16 = a₁ x (1/4)⁴
1/167 = a₁ (1/256) ==> a₁ =16 & the formula becomes
a₅ = 16(1/4)⁴
Part A;
There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.
An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.
Substituting D(-4, 2) into the system
we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.
Also, substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.
To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.
For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true
For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false
For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false
For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true
For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true
For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true
Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.
You roll a die with the sample space S = {1, 2, 3, 4, 5, 6}. You define A as {1 ,4, 6}, B as {1, 3, 4, 5, 6}, C as {1, 5}, and D
Lilit [14]
Answer:
Step-by-step explanation:
Recall that two events A,B are called mutually exclusive if and only if 
(their intersection is empty). They are exhaustive if they are mutually exclusive and their union is the sample space.
Based on this
a) Note that 
, so they are not mutually exclusive nor exhaustive.
b) 
so they are not mutually exclusive nor exhaustive.
c) 
, so they are mutually exclusive. Note that 
. Then they are exhaustive.
d) 
, so they are not mutually exclusive nor exhaustive.
