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DIA [1.3K]
2 years ago
13

A population consists of the following five values: 1, 3, 4, 4, and 6. List all samples of size 2 from left to right without rep

lacement, and compute the mean of each sample. (Round your mean value to 1 decimal place.)
Mathematics
1 answer:
aniked [119]2 years ago
8 0

Answer:

The sample of sizes 2 and their mean are given below.

Step-by-step explanation:

The population consist of  5 values, S = {1, 3, 4, 4, 6}.

The number of samples of size 2 (without replacement) that can be formed from these 5 values is:

{5\choose 2}=\frac{5!}{2!(5-2)!} =10

Th formula to compute the mean is:

\bar x=\frac{1}{n}\sum x_{i}

List the 10 samples and their mean as follows:

<u>Sample</u>                       <u>Mean</u>

(1, 3)                    \bar x=\frac{1}{2}[1+3]=\frac{4}{2}=2.0

(1, 4)                    \bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5

(1, 4)                    \bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5

(1, 6)                    \bar x=\frac{1}{2}[1+6]=\frac{7}{2}=3.5

(3, 4)                   \bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5

(3, 4)                   \bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5

(3, 6)                   \bar x=\frac{1}{2}[3+6]=\frac{9}{2}=4.5

(4, 4)                   \bar x=\frac{1}{2}[4+4]=\frac{8}{2}=4.0

(4, 6)                   \bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0

(4, 6)                   \bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0

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Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

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The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

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Determine which of (a)-(d) form a solution to the given system for any choice of the free parameter. (HINT: All parameters of a
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Answer:

  125π√3/3 cm³ ≈ 226.72 cm³

Step-by-step explanation:

The length of the circular edge of the half-circle is ...

  (1/2)C = (1/2)(2πr) = πr = 10π . . . . cm

This is the circumference of the circular edge of the cone, so the radius of the cone is found from ...

  C = 2πr

  10π = 2πr . . . . fill in the numbers; next, solve for r

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The slant height of the cone is the original radius, 10 cm, so the height of the cone from base to apex is found from the Pythagorean theorem.

  (10 cm)² = h² + r²

  h = √((10 cm)² -(5 cm)²) = 5√3 cm

And the cone's volume is ...

  V = 1/3·πr²h = (1/3)π(5 cm)²(5√3 cm)

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What is the ninth term in the binomial expansion of (X-2y)^13?
vlada-n [284]

Answer:

329472\cdot x^5\cdot y^8  

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(a+b)^n=a^n+n\cdot a^{n-1}{\cdot b}+\frac{n(n-1)}{2}\cdot a^{n-2}\cdot b^2+------+b^n

Here in given expression we have a=x , b=-2y and n = 13

general formula for binomial expansion is:

T_{r+1} =^nC_r\cdot a^{n-r}\cdot b^r

Since, T_{r+1}

r should be one number less than the term we need to find so it will become the number we need to find like here we have to find 9th term so, r=8

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After substituting the values we will get

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After simplification we will get

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x =  number of weeks required to complete the assignments

We want to find f⁻¹ (30) as an estimate of the number of weeks required to complete 30 assignments.
The procedure is as follows:

1. Set y = f(x)
   y = 4x - 3/2

2. Exchange x and y
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3. Solve for y
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4. Set y equal to f⁻¹ (x)
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5. Find f⁻¹ (30)
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Answer:
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