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Delvig [45]
3 years ago
5

They’re 80 skittles in a bag 20% of them are read how many skittles a red

Mathematics
1 answer:
Nutka1998 [239]3 years ago
7 0

Answer: 16

Step-by-step explanation: 80x20=1600

Put the decimal place where it belongs

80% of 20=16

Hope this helps!

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3 years ago
If L=√(x^2+y^2), dx/dt =-4, dy/dt=3, find dL/dt when x=4 and y=3
yulyashka [42]

Hello,

L=\sqrt{x^2+y^2} \\\\\dfrac{dx}{dt}=-4\\\\\dfrac{dy}{dt}=3\\\\x=4, y=3\\\\\dfrac{\partial L} {\partial x} =\dfrac{x}{\sqrt{x^2+y^2}} \\\\\dfrac{\partial L} {\partial y} =\dfrac{y}{\sqrt{x^2+y^2}} \\\\\dfrac{dL}{dt} =\dfrac{\partial L} {\partial x}*\dfrac{dx}{dt}+\dfrac{\partial L} {\partial y}*\dfrac{dy}{dt}\\\\=-4*\frac{-4}{\sqrt{4^2+3^2}} +3*\frac{3}{\sqrt{4^2+3^2}}\\\\=\dfrac{16}{5} +\dfrac{9}{5} \\\\=5\\

5 0
3 years ago
Pleaseeee help meee.
DiKsa [7]

Answer:

Step-by-step explanation:

13. Match it with 54

14. Match it with 576

15. Match it with 70

16. Match it with 6

Hope this helps

(Also, do you want how the answers came up? If so please comment below this and tell me that. I  would be more than happy to write the equations if you want.)

3 0
3 years ago
Triangles similar to the same triangle are ______ to each other.
wel
A is the corect answer  they congruent.
8 0
3 years ago
Read 2 more answers
Hi need help for this maths question
Sauron [17]

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}

c) The mean is given by the integral,

\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy

Integrate by parts, with

u = y \implies du = dy

dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}

Then

\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)

\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy

\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}

8 0
2 years ago
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