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2 years ago

I'll mark u as brainliest pls help with these 2 questions

Mathematics

2 answers:

son4ous [18]2 years ago

8 0

Answer:

1.) x=3

2.)x=10

Step-by-step explanation:

for the first problem you would add 1 to both sides, and you would get 5x=15, then you would divide both sides by by 5 to get x by itself. 5x/5=x and 15/5=3, so x=3. you can also plug it in. 5(3)-1=14

for the second one you wpuld subtract 15 from both sides, and you would get 3x=30. divide both sides by 3 which is x=10. plug it in, 15+3(10)=45.

aev [14]2 years ago

5 0

Answer:

3, 10

Step-by-step explanation:

5x-1=14 ->

5x=15 ->

x=3.

15+3x=45 ->

3x=45-15=30 ->

3x=30 ->

x=10.

Hope that help! :)

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3 years ago

I need to know this plz and thank u

mr_godi [17]

The third one :) I hope I did it right

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3 years ago

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Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

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3 years ago

The prices of all college textbooks follow a bell-shaped distribution with a mean of $113 and a standard deviation of $12. Using

Soloha48 [4]

Step-by-step explanation:

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68.27% of the data lies within one standard deviation of the mean.

95.45% of the data lies within two standard deviations of the mean.

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In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

$\Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})$

where the symbol $\Phi$ (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, $\mu$ is the mean and $\sigma$ is the standard deviation of the normal distribution defined as $N(\mu, \ \sigma)$ .

According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution $N(113, \ 12)$ ,

$\Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973$

Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.

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2 years ago

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yanalaym [24]

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3 years ago

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I'll mark u as brainliest pls help with these 2 questions ​

I'll mark u as brainliest pls help with these 2 questions