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Rus_ich [418]
3 years ago
7

If you earn $12 an hour and you work 40 hours a week, how much do you earn a month?

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

$1920

Step-by-step explanation:

$12/hr

40hrs a week

40×12=$480/week

as there is 4week a month

2×$480=$1920

qaws [65]3 years ago
4 0

Answer:

$1,920

Step-by-step explanation:

you multiply 12 by 40. then, assuming a typical month is 4 weeks, you multiply that number by 4.

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4-9=-5

Step-by-step explanation:

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7 0
3 years ago
Suppose you want to learn more about the heights of the buildings in New York City. How would you choose a sample of the buildin
natulia [17]

Answer:

I'd use a Non-probability Sampling Method.

Explanation:

The question is already laced with criteria - Heights of Buildings in New York.

This means that there are certain buildings that won't fit in the sample. In a non-probability sample, objects or subjects are elected based on specified criteria that are not random. This means that not all objects/subject have a c<em>hance</em> of being included in the sample.

It is assumed that the question/assignment centers around high rise buildings.

Therefore, one storey buildings and bungalows that are within and outside New York will not be included in the sample.

Under the Non-Probability Sampling Method, it is essential to note that there are other subgroupings of sampling techniques. They are:

  1. Convenience Sample
  2. Voluntary Sample
  3. Purposive Sample
  4. Snowball Sample

Cheers!

3 0
3 years ago
Read 2 more answers
d.) The expression that describes the network of logic gates: Is [A AND (NOT B)] OR (NOT B). Complete the input-output table for
marusya05 [52]
<span>Let R=A AND (NOT B)] OR (NOT B)
 If A=T, B=T
then ~B=F,
A and ~B = F
So R=F
 If A=T,
B=F ~B=T
 A and ~B=T
so (A and ~B) or ~B = T etc.</span>
3 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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