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topjm [15]
3 years ago
11

HELP ME NOW PLEASE NOW ADGAFDKD;LHJ!!!!!! PLEASE AND THANK YOU :)))))

Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0

I got D.

There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:

1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator

1 - .1131 - .2005 = .6864

Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.

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Which of the following statements is true? A. 15-0=0 B. 0÷15=0 C. 15+0=0 D. 15÷0=0​
Sergeu [11.5K]

Answer:

B) 0:15=0

Step-by-step explanation:

3 0
3 years ago
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}

4 0
3 years ago
What is the value of x in the equation below?<br><br> 7x + 8 = -13
ivann1987 [24]

Answer:

x=-3

Step-by-step explanation:

7x=-13-8

7x=--21

x=-21/7

x=-3

points or brainleist please

8 0
2 years ago
Read 2 more answers
Select True or False for each statement.
SSSSS [86.1K]

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TRUE

\sqrt[5]{36^4}=36^{4/5}

which surely isn't 36.  FALSE

\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0

FALSE

For the fourth one we have a

\sqrt{98b} + \sqrt{2b}

which isnt

10\sqrt{b}

so this is FALSE.

\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}

= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}

=(\sqrt 5 + \sqrt 6)^2

No fractions in that one so FALSE.

3 0
2 years ago
Referring to the table below, are the events is female and owns a dog independent?​
Readme [11.4K]

Answer:

No

Step-by-step explanation:

They are independent iff:

P(F & D) = P(F) × P(D)

15.278/100 = (55/100) × (27.778/100)

0.15278 = 0.125279 (false)

Therefore, not independent

3 0
3 years ago
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