In an arithmetic sequence:
Tn=t₁+(n-1)d
t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d
t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d
Therefore:
3t₁+12d=300 (1)
t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d
t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d
Therefore:
3t₁+45d=201 (2)
With the equations (1) and (2) we make an system of equations:
3t₁+12d=300
3t₁+45d=201
we can solve this system of equations by reduction method.
3t₁+12d=300
-(3t₁+45d=201)
-----------------------------
-33d=99 ⇒d=99/-33=-3
3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112
Threfore:
Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n
Now, we calculate T₁₈:
T₁₈=115-3(18)=115-54=61
Answer: T₁₈=61
2 x 3 x 2 x 3 units rectangle as shown in the picture would work!
Step-by-step explanation:
We are dealing with theoretical probability. This means that our probability of said outcome is

where x is the outcome we want, and y is the total number of outcomes possible.
28. There is 3 triangles out of 6 polygons so the probability is 1/2.
29.There is 1 Pentagon so the probability is 1/6.
30. This is a complement to question 28. A complement is the inverse of the problem. A complement and its original add to 1 so the probability of both getting a triangle is 1/2.
31. There is 4 non quadraletrial so the probability is 2/3.
32. There is 3 figures that has more sides than three so 1/2 is the probability.
33.There is only multi right angles figures so 1/2 is the probability.
34. There is 30 days in April so the probability it's the 29th is

35. There is 31 days in July and 15 days after the 16th. So the probability
is

The quantity remaining will be
.. 448*(1/2)^(24/6) = 448/16 = 28 . . . . grams