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Ronch [10]
3 years ago
12

Round to the nearest given place. 1.94155 thousandths

Mathematics
2 answers:
Scilla [17]3 years ago
7 0

Answer: 2

Step-by-step explanation: You round up when the number is 5 or higher. In this case since it’s 1.9, you can just round to 2

choli [55]3 years ago
7 0

Round off 1.94155 to thousands, when you place each of the decimal fraction into their place value you will discover that 9 is tenth, 4 is hundredth and 1 is thousandth. Since the number that follows 1 being the thousandth is 5 you will need to round up 5 to become 1 and add to the 1 in thousandth to become 2. Therefore the answer is 1.942

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B 4ab= c^2 - (b-a) ^2

Step-by-step explanation:

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Find the product of (x-7)^2 and explain how it demonstrates the closure property of multiplication
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A. x²-14x+49; is a polynomial

Step-by-step explanation:

(x-7)² can be written as (x-7)(x-7)

Expanding the expression

x(x-7)-7(x-7)

x²-7x-7x+49

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4 0
3 years ago
Identify the lower class​ limits, upper class​ limits, class​ width, class​ midpoints, and class boundaries for the given freque
maw [93]

Answer:

The number of individuals included in the summary is 146.

Step-by-step explanation:

The frequency distribution table provided is as follows:

Class Intervals    Frequency

    100 - 199               24

   200 - 299              90

   300 - 399              27

   400 - 499                1

   500 - 599               4

The lower class limit it the smallest value of each class interval.

Lower class limit = {100, 200, 300, 400, 500}

The upper class limit it the highest value of each class interval.

Upper class limit = {199, 299, 399, 499, 599}

The lower class boundaries are the lower class limits decreased by 0.5 and the upper class boundaries are the upper class limits increased by 0.5.

Class boundaries:

   99.5 - 199.5  

  199.5 - 299.5

  299.5 - 399.5

  399.5 - 499.5

  499.5 - 599.5

The class width is the difference between the class boundaries of each class.

Class width = 199.5 - 99.5 = 100

So, the class width is 100.

The midpoints of a class is the average value of the boundaries of a class.

\text{Midpoint}_{100-199}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{99.5+199.5}{2}\\\\=149.5

\text{Midpoint}_{200-299}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{199.5+299.5}{2}\\\\=249.5

\text{Midpoint}_{300-399}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{299.5+399.5}{2}\\\\=349.5

\text{Midpoint}_{400-499}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{399.5+499.5}{2}\\\\=449.5

\text{Midpoint}_{500-599}=\frac{\text{Lower class boundary} + \text{Upper class boundary}}{2}\\

                         =\frac{499.5+599.5}{2}\\\\=549.5

The number of individuals included in the summary is the sum of all frequencies.

\text{Number of Individuals}=24 + 90 + 27 + 1 + 4=146

Thus, the number of individuals included in the summary is 146.

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Answer:

They will need to hire 5 testers.

Step-by-step explanation:

This question can be solved using a simple rule of three.

For each 10 developers, we need 2.5 testers. So how many testers are needed for 20 developers?

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