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3 years ago

What is the unit rate for 0.75

Mathematics

1 answer:

kotykmax [81]3 years ago

5 0

Answer:

The unit rate for 0.75 is Answer: 75%

Step-by-step explanation:

Because it is actually converting decimals to percentage.

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the graph of a liner relationship has a slope of -5 and passes through the point (2, -12). write tge equation of the liner in th

bixtya [17]

Answer:

y = -5x - 2

Step-by-step explanation:

equation formula: y = mx + c

m = -5

-12 = -5(2) + c

c = -2

⇒ y = -5x - 2

5 0

2 years ago

Which equation could be used to find the length of the hypotenuse?

Komok [63]

Answer:

<h3>(base)² + (altitude)² = (hypotenuse) ²</h3>

Therefore,

2²+5² = c² will be matched.

7 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

8 divided by 2 (2+2). 1 or 16

laiz [17]

Answer:

Step-by-step explanation:

$8/2(2+2)$

Following the order of operations (PEMDAS), combine the terms in your parentheses:

$8/2(4)$

Remember that we follow the order of operations from left to right.

Divide 8 into 2:

$4(4)$

Multiply:

$16$

8 0

3 years ago

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally

mariarad [96]

Answer:

a) 0.1587 (or 15.87%)

b) 0.0478 (or 4.78%)

c) 0.7935 (or 79.35%)

Step-by-step explanation:

Find the following probabilities:

(a) the thickness is less than 3.0 mm

We need to find the area under the Normal curve with a mean of 4.5 mm and a standard deviation of 1.5 mm to the left of 3 mm

<h3>(See picture 1 attached) </h3>

We can do that with the help of a calculator or a spreadsheet.

<em>In Excel use </em>

<em>NORMDIST(3,4.5,1.5,1) </em>

<em>In OpenOffice Calc use </em>

<em>NORMDIST(3;4.5;1.5;1) </em>

and we get the value 0.1587 (or 15.87%)

(b) the thickness is more than 7.0 mm

Now we need the area to the right of 7 (1 - the area to the left of 7)

<h3>(See picture 2 attached) </h3>

<em>In Excel use </em>

<em>1-NORMDIST(7,4.5,1.5,1) </em>

<em>In OpenOffice Calc use </em>

<em>1-NORMDIST(7;4.5;1.5;1) </em>

and we get the value 0.0478 (or 4.78%)

(c) the thickness is between 3.0 mm and 7.0 mm

We are looking for the area between 3 and 7

<h3>(See Picture 3) </h3>

Since the area under the Normal equals 1, we have

Area to the left of 3 + Area between 3 and 7 + Area to the right of 7 = 1

Hence,

0.1587 + Area between 3 and 7 + 0.0478 = 1

and

Area between 3 and 7 = 1 - 0.1587 - 0.0478 = 0.7935 (or 79.35%)

4 0

3 years ago

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