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bagirrra123 [75]
3 years ago
5

Any one help thx I'm just stuck on this​

Mathematics
1 answer:
MArishka [77]3 years ago
5 0

Answer:

A) 9 => 6 => 3 => 0 => -3 => -6

B) Using Formula

a_{n}= a_{1}  + d(n-1)

=> a_{n } = 9+3(n-1)

=> a_{n}= 9+3n-3\\a_{n} = 3n+6

<u><em>Answer: </em></u>a_{n }= 3n+6

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I believe it is 70.65
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how to find the area of each circle. use a calculator value of pi. round the answer to the nearest tenth
tankabanditka [31]

Answer:

A = 201.0 mi^2

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

We know that r = 8

A = (3.14) *8^2

A = 3.14 * 64

A = 200.96 mi ^2

Rounding to the nearest tenth

A = 201.0 mi^2

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the pressure, p. of a gas varies inversely with its volume, v. Pressure is measured in units of Pa. Suppose that a particular am
LekaFEV [45]

I think I know how to do it


Your equation is:

P₁V₁ = P₂V₂


You Know:

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V₁ = 336 L


New:

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V₂ = 216L


Plug this into the equation:

P₁V₁ = P₂V₂

(84 Pa)(336 L) = P₂(216 L)     Divide 216 on both sides

\frac{(84Pa)(336L)}{216L}=\frac{(P_{2})(216L)}{216L}

130.6666666 = P₂    

(if your doing significant figures, you have 2 sig figs)

130 Pa = P₂


6 0
3 years ago
How do i solve this? (1/3x-11)^(1/2)=5
photoshop1234 [79]
(\frac{1}{3} x-11)^{ \frac{1}{2} }=5\\ ((\frac{1}{3} x-11)^{ \frac{1}{2} })^{2}=(5)^{2}\\ \frac{1}{3} x-11=25\\ \frac{1}{3} x=36\\x=108
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g Consider the experiment of a single roll of an honest die and a single toss of 3 fair coins. Let X be the value on the die and
klemol [59]

Answer:

The probability function of X and Y is

P(X = k, Y = 0)  = 1/48\\P(X = k, Y = 1) = 1/16\\P(X = k, Y = 2) = 1/16\\P(X = k, Y = 3) = 1/48

With k in {1,2,3,4,5,6}

Step-by-step explanation:

We can naturally assume that X and Y are independent. Because of that, P(X=a, Y=b) = P(X=a) * P(Y=b) for any a, b.

Note that, since the die is honest, then P(X=k) = 1/6 for any k in {1,2,3,4,5,6}. We can conclude as a consequence that P(X=k, Y=l) = P(Y=l)/6 for any k in {1,2,3,4,5,6}.

Y has a binomial distribution, with parameters n = 3, p = 1/2. Y has range {0,1,2,3}. Lets compute the probability mass function of Y:

P_Y(0) = {3 \choose 0} * 0.5^3 = 1/8

P_Y(1) = {3 \choose 1} * 0.5* 0.5^2 = 3/8

P_Y(2) = {3 \choose 2} * 0.5^2*0.5 = 3/8

P_Y(3) = {3 \choose 3} * 0.5^3 = 1/8

Thus, we can conclude that the joint probability function is given by the following formula

P(X = k, Y = 0) = 1/8 * 1/6 = 1/48\\P(X = k, Y = 1) = 3/8 * 1/6 = 1/16\\P(X = k, Y = 2) = 3/8 * 1/6 = 1/16\\P(X = k, Y = 3) = 1/8 * 1/6 = 1/48

For any k in {0,1,2,3,4,5,6}

4 0
3 years ago
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