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3 years ago

7

Suppose that, based on a sample, the 95% confidence interval for the mean of a population is (25, 41). What was the mean of the

sample?
A.
31
B.
37
C.
33
D.
35

1 answer:

icang [17]3 years ago

7 0

Answer:

C. 33

Step-by-step explanation:

Sample mean is right in the centre of the interval

Sample mean:

(25+41)/2

66/2

33

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A contractor records all of the bedroom areas, in square feet, of a five-bedroom house as: 100, 100, 120, 120, 180 What is the v

kifflom [539]

The variance is the total of the squared distances of the given data from the mean.

This can be calculated through the equation,

  σ² = summation of X² / N - μ²

where σ² is the variance X's are the data, N is the number of terms, and μ is the mean.

summation of X² = 100² + 100² + 120² + 120² + 180² = 81200
N = 5
  μ = (100 + 100 + 120 + 120 + 180) / 5
μ = 124

Substituting these values to the equation for variance,

  σ² = (81200/5) - 124² = 864

Thus, the variance is equal to 864.

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3 years ago

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7x + (9x + 8) = (7x + 9x) + 8 is what property

bonufazy [111]

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3 years ago

5\2,3 and 3\2,8 find the distance

pochemuha

Answer:

√26 ≈ 5.1 units

Step-by-step explanation:

√(x₂ - x₁)² + (y₂ - y₁)²

√(3/2 - 5/2)² + (8 - 3)²

√(-1)² + (5)²

√1 + 25

√26

4 0

4 years ago

51 correct 9 incorrect as a fraction

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5 and 2/3 because 51/9= 5.666666667, which .66666667 in fraction form is 2/3

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3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

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3 years ago