∑ 4 * 5^(i-1) = 4 + 20 + 100 + 500 = 624
∑ 3 * 4^(i-1) = 3 + 12 + 48 + 192 + 768 = 1,023
∑ 5* 6^(i-1) = 5 + 30 = 35
∑ 5^(i-1) = 1 + 5 + 25 + 125 = 156
Answer:
∑ (i=1, 2) 5 * 6^(i-1) < ∑ (i=1, 4) 5^(i-1) < ∑ (i=1, 4) 4 * 5^(i-1) <
< ∑ (i=1, 5) 3 * 4^(i-1)
Answer:
The p-value would be 0.0939
Step-by-step explanation:
We can set a standard t-test for the Null Hypothesis that 
The test statistic then takes the form
with this value we then can calculate the probability that is left to the right of this value 
. From theory we know that t follows a standard normal distribution. Then 
which is smaller than the p-value set by Breyers of 0.10
Answer: A (30)
Step-by-step explanation:
By defaults, data will be enabled in tens. And it increases by replicating the initial value.
There is no way the maximum number of dataflow definitions available in this situation will be 45, 25 or 35
The only possible replicant that can be available is 30
Answer:
0.2 pounds of rice
Step-by-step explanation:
To find how much rice was used by each person, divide 22 by 110
22/110
= 0.2
So, each person used 0.2 pounds of rice
Answer:
C. Kalena made a mistake in Step 3. The justification should state: -x²
+ x²
Step-by-step explanation:
Given the function x(x - 1)(x + 1) = x3 - X
To justify kelena proof
We will need to show if the two equations are equal.
Starting from the RHS with function x³-x
First we will factor out the common factor which is 'x' to have;
x(x²-1)
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;
a²-b² = (a+b)(a-b) hence;
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Factorising x(x+1) gives x²+x, therefore
x(x+1)(x-1) = (x²+x)(x-1)
(x²+x)(x-1) = x³-x²+x²-x
The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²
