The sequence is decreasing so it is r<1, therefore it is converging
This is the formula for how to find the sum/limit of the convergence (or how to find a infinite geometric sequence): a1/(1-r)
a1=48
r=8/48=.167
Verifying r:
a1/r=48*.167=8.016=8
a1/r^2=48*.167^2=1.338672=1.34
4/3=1.33
(close enough)
Putting it into equation:
a1/(1-r)=48/(1-.167)=48/.833=57.62304922
Answer Choices:
A. Converges; 288/5
B. Converges; 0
C. Diverges
D. Converges; -12432
288/5=57.6
ANSWER IS A. Converges; 288/5
Answer:
Use 4 coins. Let heads = hit and tails = miss. Toss each coin and record the results in a table. Coin 1Coin 2Coin3Coin 4Set 1HHHHSet 2HTHHSet 3HHTHSet 4THTTSet 5
Repeat the coin tosses until you have recorded 50 sets of 4 tosses each. b. Count the successful outcomes—those with three or four heads. Coin 1Coin 2Coin3Coin 4SuccessSet 1HHHHxSet 2HTHHxSet 3HHTHxSet 4THTTSet 5
Step-by-step explanation:
Answer:
The expressions in the image are:
(-1)^(1/3)
27^(1/3)
In radical form, these expressions can be written as:
The cube root of -1, which simplifies to -1.
The cube root of 27, which simplifies to 3.
So, the simplified expressions are -1 and 3 respectively.
Step-by-step explanation:
~~~~~WDFADS~~~~~
The simplifiedform of the square root of 49 is 7.
We have,
To simplify the squareroot of 49, we can break it down using the properties of square roots.
First, we can simplify the squareroot of 49, which is 7, since the square root of a perfect square is an integer.
Next, we can simplify the square root of .
We can rewrite as ()².
Then, we can take the squareroot of , which is x².
Thus,
The simplifiedform of the square root of 49 is 7.
Learn more about expressions here:
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