Let the n-th term be called

We see that if we choose

then the other numbers follow the pattern

(see :

)
Hence the sequence will be
Answer:
6 dm
Step-by-step explanation:
Triangle DBE is similar to triangle ABC, so their side lengths are proportional.
DE/AC = DB/AB
The length of DB can be found from ...
DB +AD = AB
DB = AB -AD = (15 -10) dm = 5 dm
So, we can fill in the proportion:
DE/(18 dm) = (5 dm)/(15 dm)
DE = (18 dm)·(1/3) . . . . . . . . . . simplify, multiply by 18 dm
DE = 6 dm
_____
It can be helpful to draw and label a figure.
I believe this is the complete problem.
<span>Anton bought a picnic cooler. His total bill, with tax, was $7.95. He paid 6 percent sales tax. How much did he pay for the cooler alone without the tax?
The cost of an item is 100% of its price. What number goes in place of ? in the addition problem?
</span>
The answer is the below:
<span>He paid $7.473 without tax because $7.95 x 96% or (0.96) =$7.473</span>
Answer:
<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.
Step-by-step explanation:
A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.
It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.
The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:
<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.
<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.
The significance level of the test is:
<em>α</em> = 0.05.
The test is defined as:

The decision rule for the test is:
If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.
<u>Answer:</u>
Consistent and dependent
<u>Step-by-step explanation:</u>
We are given the following equation:
1. 
2. 
3. 
For equation 1 and 3, if we take out the common factor (3 and 4 respectively) out of it then we are left with
which is the same as the equation number 2.
There is at least one set of the values for the unknowns that satisfies every equation in the system and since there is one solution for each of these equations, this system of equations is consistent and dependent.