The weight of a rock sample is measured to be 2.5 pounds. What is the percent of error in the measurement?

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

7 0

Answer:

Step-by-step explanation:

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What is an equivalent expression for 24+49

Alexus [3.1K]

Answer:

20 + 4 + 40 + 9

Step-by-step explanation:

20 + 4 is 24 and 40 + 9 is 49, so it's pretty much the same thing.

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3 years ago

"A pieces of licorice is cut into 10 equal-sized pieces. If the length of the piece of licorice is 2/3 yard, how long will each

jok3333 [9.3K]

Each piece of licorice will be 2.4 inches

Work: 2/3 yard = 2 feet = 24 inches ÷ 10 = 2.4 inches

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3 years ago

Christopher swam laps everyday in the town swimming pool. He swam 45 minutes each day, 5 days each week, for 12 weeks.

Dahasolnce [82]

45(5)=225 than multiply that by 60 because remember he only swam 5 out of 7 days a week (12x5=60) therefore 225x60= 13,500 minutes to convert to hours divide by 60 which will be 225 hours

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3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

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3 years ago

From a deck of 52 cards, one card is drawn at random. Match the following subsets with their correct probabilities.

zhenek [66]

1.3/13(if you mean jacks queens and kings)
2.1/52
3.1/2
4.1/13
5.1/4

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4 years ago

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