The hypothesis is this:
Is the chance of a coin landing on heads or tails mirted asymmetrically, or is it by chance?
Answer:
Ai. Common ratio = 2/3
Aii. First term = 54
B. Sum of the first five terms = 422/3
Step-by-step explanation:
From the question given above, the following data were obtained:
3rd term (T3) = 24
6Th term (T6) = 64/9
First term (a) =?
Common ratio (r) =?
Sum of the first five terms (S5) =?
Ai. Determination of the common ratio (r).
T3 = ar²
T3 = 24
24 = ar²....... (1)
T6 = ar⁵
T6 = 64/9
64/9 = ar⁵......... (2)
The equation are:
24 = ar²....... (1)
64/9 = ar⁵......... (2)
Divide equation 2 by equation 1.
64/9 ÷ 24 = ar⁵ / ar²
64/9 × 1/24 = r³
8/27 = r³
Take the cube root of both side
r = 3√(8/27)
r = 2/3
Thus, the common ratio is 2/3
Aii. Determination of the first term (a).
T3 = ar²
3rd term (T3) = 24
Common ratio (r) = 2/3
First term (a) =?
24 = a(2/3)²
24 = 4a/9
Cross multiply
24 × 9 = 4a
216 = 4a
Divide both side by 4
a = 216/4
a = 54
Thus, the first term (a) is 54
B. Determination of the sum of the first five terms.
Common ratio (r) = 2/3
First term (a) = 54
Number of term (n) = 5
Sum of first five terms (S5) =?
Sn = a[1 –rⁿ] / 1 – r
S5 = 54[1 – (⅔)⁵] / 1 – ⅔
S5 = 54 [1 – 32/243] / ⅓
S5 = 54 (211/243) × 3
S5 = 54 × 211/81
S5 = 6 × 211/9
S5 = 2 × 211/3
S5 = 422/3
Thus, the sum of the first five terms is 422/3
Answer:
10 boards
Step-by-step explanation:
$450-$250=$200. $20 * 10 boards = $200. $200-$200=0.
Therefore the answer is 10 boards
In order to prove the statement, we have to find a rational number x which satisfies r < x < s.
For the number: ======> x = r + s / 2,
We Have:
r = r + r / 2 < r + s / 2 < s + s /2 ======> s
It remains to prove that x is rational.
Let: ======> r = a/b and s = c/d
where: =====> a, b 6 = 0, c, and d 6 = 0 are integers.
Then: ====> x = r + s/2
= r = a/2b + s = c/2d
= ad + cb / 2bd
Therefore, where ad + cb, and 2bd 6 = 0 are integers. Therefore x is rational.
Hope that helps!!! : )