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oee [108]
3 years ago
9

Please explain this to me UwU

Mathematics
2 answers:
ki77a [65]3 years ago
7 0
Each answer possibility is a coordinate and represents a variety of an (x,y) point. The equation states that y = 32x + 27, so to find the correct answer, u must find the (x,y) coordinate that makes each side equal to the other. Plug in each number accordingly...

(y) = 32(x) + 27

a. (248) = 32(8) + 27

b. (379) = 32(11) + 27

c. (354) = 32(6) + 27

d. (288) = 32(9) + 27

now it’s just a matter of simplifying to see which equation is true.

a. 248 =/ 283

b. 379 = 379

c. 354 =/ 219

d. 288 =/ 315

The only equation that is true is B.
Mariana [72]3 years ago
7 0

Answer:

owo

Step-by-step explanation:

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Your answer would be 2.93.
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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

5 0
4 years ago
Yo i need help with 1,000,000 + - 3,000,000
Arada [10]

Answer: -2,000,000

Step-by-step explanation:

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6 0
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Read 2 more answers
A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag
ahrayia [7]

Answer:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(ZStep-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:

X \sim N(M,4)  

Where \mu=M and \sigma=4

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(Z

8 0
3 years ago
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