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ss7ja [257]
3 years ago
15

Biologists conducted a study to investigate the flying velocity of mosquitoes both before and after feeding. The following scatt

erplot shows the velocity after feeding, in centimeters per second, and the proportional increase in weight after feeding relative to the weight before feeding. For example, 0.5 indicates a 50 percent weight gain after feeding. One point on the graph is labeled M.
Mathematics
1 answer:
Lesechka [4]3 years ago
4 0

Answer:

It represents a mosquito that flew very fast after feeding relative to all other mosquitoes.

Step-by-step explanation:

The point is unusual because the velocity of the mosquito is substantially greater than all other velocities and is probably an outlier.

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Nico rolled two numbers 250 times
sleet_krkn [62]

Answer:

Probability is O %

Step-by-step explanation:

3 0
3 years ago
Factorise completely:<br> 1. 3xy - x<br>2. 4a² - 9​
Amanda [17]

Answer:

x(3y - 1) and (2a - 3)(2a + 3)

Step-by-step explanation:

(1)

3xy - x ← factor out x from each term

= x(3y - 1)

(2)

4a² - 9 ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

4a² - 9

= (2a)² - 3²

= (2a - 3)(2a + 3)

6 0
2 years ago
Help please ........
Nitella [24]

Answer:

D

Step-by-step explanation:

eh too lazy to explain but plz trust me

5 0
3 years ago
T=m-n for n for the final answer
Snezhnost [94]

Answer:n=-t+m

Step-by-step explanation:

6 0
2 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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