Theos initial balance in his savings account = -$4
Final balance after making a deposit = $25
So, the overall change to his account = $ 29
Answer:
the common ratio is either 2 or -2.
the sum of the first 7 terms is then either 765 or 255
Step-by-step explanation:
a geometric sequence or series of progression (these are the most common names for the same thing) means that every new term of the sequence is created by multiplying the previous term by a constant factor which is called the common ratio.
so,
a1
a2 = a1×f
a3 = a2×f = a1×f²
a4 = a3×f = a1×f³
the problem description here tells us
a3 = 4×a1
and from above we know a3 = a1×f².
so, f² = 4
and therefore the common ratio = f = 2 or -2 (we need to keep that in mind).
again, the problem description tells us
a2 + a4 = 30
a1×f + a1×f³ = 30
for f = 2
a1×2 + a1×2³ = 30
2a1 + 8a1 = 30
10a1 = 30
a1 = 3
for f = -2
a1×-2 + a1×(-2)³ = 30
-10a1 = 30
a1 = -3
the sum of the first n terms of a geometric sequence is
sn = a1×(1 - f^(n+1))/(1-f) for f <>1
so, for f = 2
s7 = 3×(1 - 2⁸)/(1-2) = 3×-255/-1 = 3×255 = 765
for f = -2
s7 = -3×(1 - (-2)⁸)/(1 - -2) = -3×(1-256)/3 = -3×-255/3 =
= -1×-255 = 255
Answer:
6
Step-by-step explanation:
I wrote it super fast. lol
can you get it?
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
Just replace a with 4 and b with 2 and your equation looks like
3.14 (4×2+4×2)
solve within the parenthesis first
3.14 (8+8)
3.14 (16) =50.24
