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MariettaO [177]

3 years ago

8

g In cross-tabulation, one of the tests used to evaluate the statistical association between two categorical or ordinal variable

s is called: The Chi-square test of independence The product-moment correlation coefficient The test of mean equality for independent samples The parametric test of dependence

1 answer:

Lena [83]3 years ago

5 0

Answer:

The Chi-square test of independence

Step-by-step explanation:

The answer to this question is the Chi-square test of independence because it is what is used to determine the existence of an association between nominal/categorical variables. This test is non parametric in nature and it can also be called the chi square test of association.

The data is tabulated. each row is a representation of the category for one of the variables and each column is for the other variable.

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Estimate the value of 196 divided by 0.499

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Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a

Vitek1552 [10]

Answer:

$2ab - 6a + 5b - 15$

Step-by-step explanation:

Given

$2ab - 6a + 5b + \_$

Required

Fill in the gap to produce the product of linear expressions

$2ab - 6a + 5b + \_$

Split to 2

$(2ab - 6a) + (5b + \_)$

Factorize the first bracket

$2a(b - 3) + (5b + \_)$

Represent the _ with X

$2a(b - 3) + (5b + X)$

Factorize the second bracket

$2a(b - 3) + 5(b + \frac{X}{5})$

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$b - 3 = b + \frac{X}{5}$

Subtract b from both sides

$b - b- 3 = b - b+ \frac{X}{5}$

$- 3 = \frac{X}{5}$

Multiply both sides by 5

$- 3 * 5 = \frac{X}{5} * 5$

$X = -15$

Substitute -15 for X in $2a(b - 3) + 5(b + \frac{X}{5})$

$2a(b - 3) + 5(b + \frac{-15}{5})$

$2a(b - 3) + 5(b - \frac{15}{5})$

$2a(b - 3) + 5(b - 3)$

$(2a + 5)(b - 3)$

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3 years ago

Wilmer went up a hill for x minutes at a speed of y kilometers per minutes. Then he went down the same path at a speed of z kilo

dedylja [7]

Answer:

$xy=wz$

Step-by-step explanation:

Given that:

Speed of Wilmer up the hill = $y$ km/min

Time taken up the hill = $x$ minutes

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Distance can be calculated by multiplying the Speed with Time i.e.

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It is well known that, the distance up the hill and down the hill will be same.

Putting both the values equal, we get:

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