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3 years ago

Hello.Please help me on number 12

Mathematics

2 answers:

vampirchik [111]3 years ago

8 0

This problem is asking for a ration between soccer balls and footballs.

All that you need to do is count the number of both balls,

6:10.

Charra [1.4K]3 years ago

7 0

There are 6 soccer balls and 10 footballs
the ratio of soccer balls to footballs is 6:10
to simplify that you divide 2 by each and its simplified to 3:5

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What is the length of the radius of a circle with a center at 2 + 3i and a point on the circle at 7 + 2i?

dem82 [27]

You can calculate the difference/vector of both points:
$7+2i-(2+3i)=5-i$
then calculate the magnitude/length like pythagoras
$radius=\sqrt{5^2+(-i*-i)}\\=\sqrt{5^2+i^2}\\=\sqrt{5^2-1}\\=\sqrt{24}\\=\sqrt{4*6}\\=2*\sqrt{6}$

7 0

3 years ago

Read 2 more answers

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on three numbers pays 11 to 1 (that is, if you bet $1 and one of

OLga [1]

Answer:

You are expected to lose $0.05 (or win -$0.05)

Step-by-step explanation:

Since the roulette wheel has the numbers 1 through 36, 0, and 00, there are 38 possible outcomes.

In this bet, you are allowed to pick 3 out of the 38 numbers. Thus, your chances of winning (P(W)) and losing (P(L)) are:

$P(W)=\frac{3}{38}\\P(L) = 1 - P(W)\\P(L) = \frac{35}{38}\\$

The expected value of the bet is given by the sum of the product of each outcome pay by its probability. Winning the bet means winning $11 while losing the bet means losing $1. The expected value is:

$EV = (11*\frac{3}{38}) -(1*\frac{35}{38})\\EV = -\$0.0525$

Therefore, with a $1 bet, you are expected to lose roughly $0.05

5 0

3 years ago

koda Buys 0.75 kg of cortos,which ia 5times the mas of the union he also Buys. How much does the union weigh?

kari74 [83]

Answer:

0.15kg

Step-by-step explanation:

Given data

We are told that

0.75 kg of cortos weights 5times the mas of the onion

We want to find the mass of 1 onion

Hence

0.75 kgcortos = 5 onions

x cortos = 1 onions

x= 0.75/5

x= 0.15kg

Hence 1 onion will weigh 0.15kg

4 0

3 years ago

use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.

Each slice is a quarter of cylinder 5/n thick and has a radius of

-k(5/n) + 5 for each k = 1,2,..., n (see picture)

So the volume of each slice is

$\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}$

for k=1,2,..., n

We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

Notice that the last term of the sum vanishes. After making up the expression a little, we get

$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

we also know that

$\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}$

and

$\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}$

so we have, after replacing and simplifying, the sum of the slices equals

$\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}$

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$