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antiseptic1488 [7]
3 years ago
13

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Mathematics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

Class Boundary = 1 between the sixth and seventh classes.

Step-by-step explanation:

              Lengths (mm)                   Frequency

1.                  140 - 143                                  1

2.                 144 - 147                                 16

3.                 148 - 151                                 71

4.                 152 - 155                              108

5.                 156 - 159                               83

6.                 160 - 163                                18

7.                  164 - 167                                 3

The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.

Therefore the class boundary between the sixth and seventh classes

= 164 - 163  = 1

Therefore Class Boundary = 1.

It can be seen that class boundary for the frequency distribution is 1.

If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.

Therefore, Class width,

w = lower limit of second class - lower limit of first class

   = 144 - 140

   = 4

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An agency has 4,000 employees. 2,000 employees have at least two years of college education. 1,000 of those employees have bache
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Step-by-step explanation:

Given that:

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3 years ago
Hello, can you explain this topic im in 6th grade ​
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Hello there I hope you are having a great day :)

<u>The Topic Shifting digits:</u>

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3 years ago
Read 2 more answers
Could someone help me for this?
lina2011 [118]

Answer:

x = - \frac{5}{3} , x = \frac{5}{2}

Step-by-step explanation:

to find the points of intersection equate the 2 equations , that is

7x - 15 = 10 + 12x - 6x² ( subtract 10 + 12x - 6x² from both sides )

6x² - 5x - 25 = 0 ← factor the quadratic on left side

consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 6 × - 25 = - 150 and sum = - 5

the factors are - 15 and + 10

use these factors to split the x- term

6x² - 15x + 10x - 25 = 0 ( factor the first/second and third/fourth terms )

3x(2x - 5) + 5(2x - 5) = 0 ← factor out (2x - 5) from each term

(2x - 5)(3x + 5) = 0

equate each factor to zero and solve for x

3x + 5 = 0 ⇒ 3x = - 5 ⇒ x = - \frac{5}{3}

2x - 5 = 0 ⇒ 2x = 5 ⇒ x = \frac{5}{2}

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