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anygoal [31]
3 years ago
9

Ingrid dug a trench 18/20 of a meter long. The next day she dug 9/20 of a meter more of the trench. What is a reasonable estimat

e of the total length of the trench?​
Mathematics
1 answer:
lutik1710 [3]3 years ago
6 0
Omg sorry i don’t really know i hope someone answers your question
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vaieri [72.5K]
Is is literally 
19/100
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Elijah's workout at the gym consists of just the elliptical and treadmill. After burning 450 miles
Galina-37 [17]

Answer:

375

Step-by-step explanation:

12.5=1min

30min=?

30x12.5=357

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Miranda has earned scores of 20, 22, 18, and 24 on her last four math quizzes. What score will she need to earn on the fifth qui
BigorU [14]

Answer:

16

Step-by-step explanation:

20 + 22 + 18 + 24 + 16 = 100

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hopes this helps sorry for the late answer I just want others to get the correct answer.

4 0
3 years ago
he​ half-life for a link on a social network website is 2.6 hours. Write an exponential function T that gives the percentage of
pychu [463]

Answer:

% Remaining= [1-(1/2)^{\frac{t}{2.6}}]x100

And replacing the value t =5.5 hours we got:

% Remaining= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%

Step-by-step explanation:

Previous concepts

The half-life is defined "as the amount of time it takes a given quantity to decrease to half of its initial value. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not".

Solution to the problem

The half life model is given by the following expression:

A(t) = A_o (1/2)^{\frac{t}{h}}

Where A(t) represent the amount after t hours.

A_o represent the initial amount

t the number of hours

h=2.6 hours the half life

And we want to estimate the % after 5.5 hours. On this case we can begin finding the amount after 5.5 hours like this:

A(5.5) = A_o (1/2)^{\frac{5.5}{2.6}}

Now in order to find the percentage relative to the initial amount w can use the definition of relative change like this:

% Remaining = \frac{|A_o - A_o(1/2)^{\frac{5.5}{2.6}}|}{A_o} x100

We can take common factor A_o and we got:

% Remaining= [1-(1/2)^{\frac{t}{2.6}}]x100

And replacing the value t =5.5 hours we got:

% Remaining = [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%

5 0
3 years ago
Liz and Sara are making cookies for a bake sale. The graph shows the number of cookies that Liz bakes during a 5-hour period.
Viktor [21]
Liz would end up baking more cookies than sara. So the answer would be d
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3 years ago
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