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2 years ago

HELP! BRainliest +20

Mathematics

1 answer:

White raven [17]2 years ago

5 0

The value of f(32) is 3

<h3>What is a Function ?</h3>

A function is a mathematical statement used to relate two variables .

It is given that

8f(x) = f((1/8)x)

It is also given that

f(4) = 24

now to determine the value of f(32)

8 * f(32) = f((1/8) *32)

8 * f(32) = f(4 )

f(32) = 24 /8 = 3

Therefore the value of f(32) is 3

To know more about function

brainly.com/question/12431044

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HELP RN RN RN RN PLEASE ASAP

lapo4ka [179]

I think the answer is A) stratified random sampling! Stratified random sampling is when sunsets of individuals are created based on similar criteria, which sounds the closest to the problem because stratified can split a group and does not have to be fully equal.
Non random sampling doesn’t fit because it’s clearly stated that it’s random.
Systematic random sampling is based on intervals in a group.
The next closest answer would be simple random, which is when a subset of individuals are chosen from a larger group with all having the same probability.

3 0

2 years ago

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .