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Inessa05 [86]
3 years ago
10

Pls help me ASAP show work please

Mathematics
1 answer:
wariber [46]3 years ago
8 0

Answer:

$3.48 for each pair of socks.

Step-by-step explanation:

It's a little confusing, because it says they're buying 2 socks, but then asks for the price of each of 2 *pairs* of socks. I'm going to believe the first sentence should say they paid for 2 pairs of socks.

They have $10, and get $3.04 in change. So we'll subtract $3.04 from $10 to find out the cost of 2 pairs of socks.

10 - 3.04 = 6.96

Then divide $6.96 by 2 pairs of socks

6.96/2 = $3.48 for each pair of socks.

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alex41 [277]

Step-by-step explanation:

there 5^(n+1) + 5^(n+2) = 5^n x 5^1 + 5^n x 5^2 breaking them as x^(a+b) = x^a x x^b

then taking common 5^n from both terms

5 0
3 years ago
200 miles divided by 8​
yulyashka [42]

Answer:

25

200 divided by 8 is 25.

Step-by-step explanation:

6 0
3 years ago
How to find the average squared distance between the points of the unit disk and the point (1,1)
ddd [48]
The unit disk can be parameterized by the function

\mathbf p(r,\theta)=(r\cos\theta,r\sin\theta)

where 0\le r\le 1 and 0\le\theta\le2\pi. The squared distance between any point in this region (x,y)=(r\cos\theta,r\sin\theta) and the point (1, 1) is

(x-1)^2+(y-1)^2=(r\cos\theta-1)^2+(r\sin\theta-1)^2
=(r^2\cos^2\theta-2r\cos\theta+1)+(r^2\sin^2\theta-2r\sin\theta+1)
=r^2(\cos^2\theta+\sin^2\theta)-2r(\cos\theta-\sin\theta)+2
=r^2-2r(\cos\theta-\sin\theta)+2

The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.

\dfrac{\displaystyle\iint_{x^2+y^2
=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}
=\dfrac{\frac{5\pi}2}\pi=\dfrac52
5 0
3 years ago
I need to be walked through this please.
Aleksandr [31]
Https://us-static.z-dn.net/files/dd1/572d05be5373c1dc9c067ca6690a41a1.jpeg

3 0
3 years ago
A stick is broken into two pieces, at a uniformly random break point. Find the CDF and average of the length of the longer piece
Oksanka [162]

Answer:

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

Step-by-step explanation:

let assume that stick has length 1.Random variable L that make length of a longer piece and random variable U that mark point .See that L < l means that

U≤ l and 1-U ≤l

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

this means 1-l≤U≤l

so we have

if we have L  [1/2,1]

then apply the formula we have E(L)=3/4

5 0
3 years ago
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