MATHEMATICS COLLEGE

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below. Lengths (mm) Frequency
140 - 143 1
144 - 147 16
148 - 151 71
152 - 155 108
156 - 159 83
160 - 163 18
164 - 167 3

What is the class boundary between the sixth and seventh classes?

Answers

Answer 1
Answer:

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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Answers

Answer:

Step-by-step explanation:

hello friend

the answer is (x)/(3) + (2)/(3)

To find the inverse, interchange the variables and solve for y

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A vehicle designed to operate on a drag strip accelerates from zero to 30m/s while undergoing a straight line path displacement of 45m. What is the vehicle's acceleration if its value may b assumed to be constant? A.2.0m/s
B.5.0m/s
C.10m/s
D.15m/s​

Answers

The vehicle's acceleration if its value may b assumed to be constant is 10 m/s. Therefore, option C is the correct answer.

Given that, a vehicle designed to operate on a drag strip accelerates from zero to 30m/s while undergoing a straight line path displacement of 45m.

What is an acceleration?

Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction-or change both.

Here, Δx = 45 m, v₀ = 0 m/s and v = 30 m/s

Using the formula v² = v₀² + 2aΔx

(30 m/s)² = (0 m/s)² + 2b (45 m)

b = 10 m/s

The vehicle's acceleration if its value may b assumed to be constant is 10 m/s. Therefore, option C is the correct answer.

Learn more about the acceleration here:

brainly.com/question/12550364.

#SPJ2

Answer:

C. 10 m/s²

Step-by-step explanation:

Given:

Δx = 45 m

v₀ = 0 m/s

v = 30 m/s

Find: a

v² = v₀² + 2aΔx

(30 m/s)² = (0 m/s)² + 2a (45 m)

a = 10 m/s²

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Is {(98,6),(99,0),(100,7)} a function?

Answers

Answer:

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An investment project involves an immediate outlay of $8 million. The net cash flows received at the end of years 1, 2, and 3 will be $3 million, $4 million, and $2 million. A 10% discount rate is applicable so that the present value factors for years 1, 2, and 3 are 0.9091, 0.8264, and 0.7513. The NPV of the investment will be: (a) (b) (c) (d) $1 million $9 million $0.46 million $7.54 million

Answers

Answer:

(c) 0.46 million

Step-by-step explanation:

As provided immediate cash outlay = $8 million.

This will represent cash outflow at period 0, as it is made immediately, no time period has lapsed.

Cash inflows as provided and the respective present value factor are:

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1                 $3 million         0.9091                $2,727,300

2                $4 million         0.8264               $3,305,600

3                 $2 million         0.7513                $1,502,600

Total present value of cash inflow = $7,535,500

Therefore, net present value = $7,535,500 - $8,000,000 = - $464,500

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Correct option is

(c) 0.46 million
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