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Sever21 [200]
4 years ago
8

Help me please with math

Mathematics
1 answer:
fomenos4 years ago
4 0

Let's make a table of differences

0 3 8 15 24

3  5 7  9

 2  2  2

Constant second difference means a quadratic polynomial,

f(n)=an² + bn +c

f(1)=0, f(2)=3, f(3)=8

0 = a+b+c

3 = 4a + 2b + c

8 = 9a + 3b + c

Subtracting pairs

3 = 3a+b

5 = 5a + b

Subtracting,

2 = 2a

a = 1

b = 5 - 5a = 0

c = -a-b = -1

f(n)=n² -1

Duh, shoulda noticed that at the beginning.

f(17)=17²-1 = 288

a₁ =5

aₙ = aₙ₋₁ - 7

aₙ = a₁ + d(n-1)

aₙ = 5 - 7(n-1)

a₇ = 5 - 7(6) = -37

Answer: -37

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Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the
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Answer:

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Step-by-step explanation:

1) Some definitions

The standard normal distribution is a particular case of the normal distribution. The parameters for this distribution are: the mean is zero and the standard deviation of one. The random variable for this distribution is called Z score or Z value.

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Based on this definition and analyzing the question :"Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value?".

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So for this case the correct function to use is: NORM.S.INV(0.975)

And the result after use this function is 1.96. And we can check the answer if we look the picture attached.

6 0
4 years ago
If the following system of equations was written as a matrix equation in the form AX = C, and matrix A was expressed in the form
IrinaVladis [17]

Answer: a-b+c+d =4


Step-by-step explanation:

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2x+8y=7\\4x-2y=9

from this we have the following matrices

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the given matrix A =\begin{bmatrix}\\a &c \\ \\b &d \\\end{bmatrix}

On comparing Matrix  A_1 with Matrix A

\begin{bmatrix}\\a &c \\ \\b &d \\\end{bmatrix}=\begin{bmatrix}\\2&8 \\ \\4 &-2 \\\end{bmatrix}

we have the following values

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8 0
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Answer:

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Step-by-step explanation:

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__

The applicable rules of logarithms are ...

  log(ab/c) = log(a) +log(b) -log(c)

  log(a^b) = b·log(a)

6 0
3 years ago
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