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Two independent random variables X and Y with various 1 and 2

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Probability (only first two tosses are heads) P( H,H,T,T,T T )

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Here all the values are given. We just need to incorporate the values in the formula which will be

Value of t = (290 – 310) / (50 / √16)

T Value = -1.60

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Number of permutations with ‘2’ in the first position =19!

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Mean = (9+13+5+9+6+5+9)/7 = 56/7 = 8

Mode = Mode is that the most repeated value of the given data set. = 9 (repeated 3 times within the set of data)

For median, first, we want to rearrange the worth in ascending order within the given data set: 5,5,6,9,9,9,13. Here, the amount 9 is placed within the middle. Hence, 9 is that the median for the given data set. So, the worth of Mean, mode, and median are 8,9,9.

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Let n1 and n2 be the number of observations in two groups having means x̄1 and x̄2, respectively. Then,

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Population mean = μ = 4.91

Population standard deviation= σ = 0.72

Sample size = n = 20 (which is less than 30)

Since the sample size is smaller than 30, use the t-score instead of the z-score, even though the population standard deviation is known.

Now, find the t-score:

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To find the Taylor Series for a function we will need to determine a general formula for

we already have a Taylor Series for then,

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First, use property 2 to divide the limit into three separate limits. Then use property to bring the constants out of the first two. This gives,

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Compute the function’s derivative, f′(x):

Find all critical numbers c such that either (i) f′(x)=0 or (ii) f’(x) is undefined.

Compute the values of f(a) and f(b) to see how they compare to the values of f(c)

has its global maximum at x = 0.1

and its global minimum at x = 1

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We have to decide whether we can find real numbers r, s,t, which are not all zero, such that rp(x) + sq(x) + tr(x) = 0.

That is:

0 = r(1 + 3x + 2x^2 ) + s(3 + x + 2x^2 ) + t(2x + x^2 )

= (r+3s) + (3r+s+2t)x+ (2r+2s+t)x2 .

This corresponds to solving the following system of linear equations

r +3s = 0

3r +s +2t = 0

2r +2s +t = 0

We compute:

Hence, {p(x), q(x), r(x)} is linearly dependent.

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The matrix A is a 3 × 3 matrix, so it has 3 eigenvalues in total. The eigenspace E₇ contains the vectors (1, 2, 1)T and (1, 1, 0)T , which are linearly independent. So E₇ must have dimension at least 2, which implies that the eigenvalue 7 has multiplicity at least 2.

Let the other eigenvalue be λ, then from the trace λ+7+7 = 2, so λ = −12. So the three eigenvalues are 7, 7 and -12. Hence, the determinant of A is 7 × 7 × −12 = −588.

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The given matrix has only one 3rd-order minor. In order that the rank arrives at 2, we must bring its determinant to zero. Hence, by applying the invariance method we can obtain values of x.

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The given equation can be written in a matrix form as AX = D and then by obtaining A–1 and multiplying it on both sides we can solve the given problem.

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