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Anika [276]
3 years ago
15

The exact same experiment was conducted 14 times. How many times should the results have been similar for them to be valid?A.8

B.5C.14D.7​
Mathematics
1 answer:
NeTakaya3 years ago
4 0

Answer:

I would have to say C.14

Step-by-step explanation:

Because if you are running the same exact emperiment with the same exact materials everytime then you should get the same outcome.

Hope this helps! : D

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The Extreme Rock Climbing Club planned a climbing expedition. The total cost was $⁢1800, which was to be divided equally among t
CaHeK987 [17]
Let's say "p" people were going to the expedition initially, and the cost for each was "c", now, we know the total cost is 1800, so for "p",  folks that'd be  1800/p how much each one cost, namely, how many times "p" goes into 1800.

well, prior to leaving, 15 dropped out, so that leaves us with " p - 15 ", and the cost "c" bumped up to " c + 27 " for each.

\bf \begin{cases}
\cfrac{1800}{p}=\boxed{c}\\\\
\cfrac{1800}{p-15}=c+27\\\\
----------\\
\cfrac{1800}{p-15}=\boxed{\cfrac{1800}{p}}+27
\end{cases}\\\\
-------------------------------\\\\
\cfrac{1800}{p-15}={\cfrac{1800}{p}}+27\impliedby 
\begin{array}{llll}
\textit{let's multiply both sides by}\quad  p(p-15)\\
\textit{to get rid of the denominators}
\end{array}

\bf 1800p=1800(p-15)+27[p(p-15)]
\\\\\\
1800p=1800p-27000+27(p^2-15p)
\\\\\\
0=-27000+27(p^2-15p)\implies 0=-27000+27p^2-405p
\\\\\\
\textit{now, let's take a common factor of }27
\\\\\\
0=p^2-15p-1000\implies 0=(p-40)(p+25)\implies p=
\begin{cases}
\boxed{40}\\
-25
\end{cases}

well, you can't have a negative value of people... so it has to be 40.

so, 40 folks were initially going, then 15 dropped out, how many went on the expedition?  40 - 15.
3 0
4 years ago
The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in th
Step2247 [10]

Answer:

\frac{dP}{dt} \ =\  \frac{1}{5}P

Step-by-step explanation:

Given

Variation: Directly Proportional

Insects = 10 million when Rate = 2 million

i.e. P = 10\ million when \frac{dP}{dt} = 2\ million

Required

Determine the differential equation for the scenario

The variation can be represented as:

\frac{dP}{dt} \ \alpha\ P

Which means that the rate at which the insects increase with time is directly proportional to the number of insects

Convert to equation

\frac{dP}{dt} \ =\ k * P

\frac{dP}{dt} \ =\ kP

Substitute values for \frac{dP}{dt} and P

2\ million = k * 10\ million

Make k the subject

k = \frac{2\ million}{10\ million}

k = \frac{2}{10}

k = \frac{1}{5}

Substitute \frac{1}{5} for k in \frac{dP}{dt} \ =\ k * P

\frac{dP}{dt} \ =\  \frac{1}{5}* P

\frac{dP}{dt} \ =\  \frac{1}{5}P

8 0
3 years ago
Emma took 35 pictures of flowers on a nature walk and an additional 3 pictures of each bird she saw. She took a total of 83 pict
AnnZ [28]
I think the last one, but not positive
6 0
4 years ago
Read 2 more answers
Let f(x)=4+3x and g(x)=x^2-3. find each function value: (f+g)(-4)
dem82 [27]

Answer:

43

Step-by-step explanation:

You are going to want to substitute the x of the function f(x) with the function of g(x) and replace the x in g(x) with -4

(f+g)(-4) = 4+3((-4)^2-3)

=43

hope this helped!

4 0
3 years ago
Read 2 more answers
Plzzzzzzzzzzzzzzzzzzzz help
shutvik [7]
(b
if you flip up the squares on the right you can see it creates the perfect right triangle.
8 0
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