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Natasha2012 [34]
2 years ago
12

A scientist wants to find the average mass of fish living in a large pond. So she will use the masses of a sample of fish from t

he pond to find the average mass.
Answer the questions below.

Mathematics
1 answer:
Oxana [17]2 years ago
6 0
The scientist would use milligrams, grams, and kilograms.

The best procedure would be to randomly pick 52 fish from the entire pond and weigh each of them as this would be the most unbiased way to achieve the average mass
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Please help me I need this answer
alexandr402 [8]

Answer:

Then answer is 10.

Step-by-step explanation:

9 would cover 1,800 square feet so 925 square feet would be left so you should buy 10 cans.

8 0
2 years ago
Read 2 more answers
What’s the length of three pipes, when combined they equal 72m, how do I find the answer
allochka39001 [22]

Answer:

Each pipe is 24m

Step-by-step explanation:

<u>Given:</u>

  • three pipes
  • combined equal 72m
  • let the variable x represent the length of 1 pipe

<u>Algebraic Expression</u>:

3 pipes times length equals 72m (the total length)

3x = 72, we want x to be alone so divided both sides by 3

3x / 3 = 72 / 3

x = 24

Learn more about Algebraic Expression here: brainly.com/question/4344214

7 0
2 years ago
A-2b= -10 for b plsssss helpppp
Artyom0805 [142]
A-2b=-10
b=-10-a.
Hope this is the answer
Mark as brainliest
5 0
3 years ago
An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d
denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. w\times 2w\times h=36

h=\frac{18}{w^2}

The equation form when top is open,

f(w)=2w^2+2wh+2(2w)h

Substitute the value of h,

f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})

f(w)=2w^2+\frac{36}{w}+\frac{72}{w}

f(w)=2w^2+\frac{108}{w}

Derivate w.r.t 'w',

f'(w)=4w-\frac{108}{w^2}

For critical point put it to zero,

4w-\frac{108}{w^2}=0

4w=\frac{108}{w^2}

w^3=27

w^3=3^3

w=3

Derivate the function again w.r.t 'w',

f''(w)=4+\frac{216}{w^3}

For w=3, f''(3)=4+\frac{216}{3^3}=12 >0

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = \frac{18}{3^2}=2\ ft

Therefore, the dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

4 0
3 years ago
PLEASE HELP
Rom4ik [11]

The answer is 15 square ft

7 0
3 years ago
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