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2 years ago

What’s the absolute value of 3

Mathematics

1 answer:

RSB [31]2 years ago

4 0

Answer: 3

Explanation: Absolute value means <em>distance from zero</em> on a number line. So far the absolute value of 3, we know that 3 is 3 units from zero on a number line so the absolute value of 3 is 3.

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SOMEBODY PLEASE HELP!!!!<br> GIVING 98 POINTS

marusya05 [52]

What is the question if it 98-49 than it would be 49

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2 years ago

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the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

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3 years ago

The ratio of green marbles to yellow marbles in Toby’s bag is equal to 2:3. What percent of the marbles in the bag are green mar

Andrews [41]

40% is the answer. you create a ratio and percent table is a way of solving it. Hope this helps!

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2 years ago

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PLZ I NEED HELP!!! THIS IS TIMED!!!! PLZ HELP ME!

wel

Answer:

it is b

Step-by-step explanation:

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Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

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2 years ago