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lys-0071 [83]
3 years ago
11

You are considering a 5/1 ARM. What does the 1 represent?

Mathematics
2 answers:
Lerok [7]3 years ago
8 0

Answer:

  B. The number of years between adjustments in the interest rate

Step-by-step explanation:

The "5" in "5/1" means there will not be any adjustment in interest rate for the first 5 years.

The "1" in "5/1" means interest rates may be adjusted at 1-year intervals after the initial 5-year period.

svp [43]3 years ago
4 0

Answer: The number of years between adjustments in the interest rate

Step-by-step explanation:

APEX

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Scientists studying biodiversity of amphibians in a rain forest have discovered that the number of species is decreasing by 2% p
kupik [55]

Answer:

Select the correct answer.

Scientists studying biodiversity of amphibians in a rain forest have discovered that the number of species is decreasing by 2% per year. There are currently 74 species of amphibians in the rain forest. Which logarithmic function models the time, , in years, it will take the number of species to decrease to a value of n?

The answer is D.

Step-by-step explanation:

8 0
2 years ago
A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

5 0
2 years ago
I need help what option
uysha [10]

Answer:

24 es la b espero que te sirva

7 0
3 years ago
Round 7.954 to the place named
Andrej [43]
What is it rounded too?
4 0
2 years ago
Solve for x. one fifth times the absolute value of the quantity x minus 4 end quantity minus 3 equals 6
Natasha_Volkova [10]

By solving an absolute value equation, we will see that the two solutions are:

x = 49 and x = -41

<h3>How to solve the equation for x?</h3>

First, we need to write the equation, we will get:

(1/5)*|x - 4| - 3 = 6

That is an absolute value equation, first let's isolate the absolute value part on the left side, so we get:

(1/5)*|x - 4| = 6+ 3 = 9

|x - 4| = 9*5 = 45

Now we have the simpler equation:

|x - 4| = 45

Which can be written in two ways:

x - 4 = 45

x - 4 = -45

So we will have two solutions for x:

x = 45 + 4 = 49

x = -45 + 4 = -41

These are the two possible values of x.

If you want to learn more about absolute value equations:

brainly.com/question/5012769

#SPJ1

5 0
2 years ago
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