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evablogger [386]
3 years ago
5

Kai had a gross weekly paycheck of $616 last week. Kai worked 6 hours for 4 of the days and 8 hours on 1 day. What is Kai's hour

ly rate of pay? a. $16.21 b. $19.25 c. $20.53 d. $25.67 Please select the best answer from the choices provided
Mathematics
2 answers:
Lisa [10]3 years ago
5 0

Answer:

B

Step-by-step explanation:

gizmo_the_mogwai [7]3 years ago
4 0

The correct answer is B. $ 19.25

Explanation:

To calculate the hourly rate of pay, first, let's calculate the total number of hours Kai worked, and then divide the total earned into the number of hours.

6 hours x 4 days = 24 hours

8 hours x 1 day = 8 hours

24 hours + 8 hours = 32 hours

This shows Kai worked a total of 32 hours. Now to find the hourly rate of pay, follow this procedure:

$616 (total earned) ÷ 32 hours = $19.25 each hour

This means Kai earns $19.25 for each hour of work and therefore the hourly rate of pay is $19.25.

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Scott's gross pay is $2344. How much of his gross pay will be deducted for the Social Security Tax?
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Answer:


Step-by-step explanation:

an employee is taxed 6.2 percent of their gross pay for Social Security, up to $127,200. After an employee has earned that much, Social Security payments are no longer deducted for the remainder of the year.

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A church has 8 bells in its bell tower. Before each church service 5 bells are rung in sequence. No bell is rung more than once.
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Use the ! tool to find the # of combinations.

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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 elementsa,\, 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.

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3 years ago
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mr Goodwill [35]

Answer:

28.64862

Step-by-step explanation:

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Please help!! also please explain I would like to understand.​
ziro4ka [17]

Answer:

A

Step-by-step explanation:

First, put this equation in slope intercept form.

y=mx+b

Where m is the slope and b is the y intercept.

y+1>2(x-1) \\ \\ y+1>2x-2 \\ \\ y>2x-3

Since it's greater than, the solution set is above the line, effectively eliminating graph D.

Now we should graph it.

We can see that the point (2,1) is on the graph, and only Graph A has that point. Therefore, Graph A is the correct graph.

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