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3 years ago

How do you do 20 / 4/5

Mathematics

2 answers:

Flauer [41]3 years ago

4 0

What do you mean by this

tamaranim1 [39]3 years ago

4 0

Answer:

Step-by-step explanation:

you can write this equation as 20 ÷ $\frac{4}{5}$

so change the division sign to multiplication by flipping the divisor ( $\frac{4}{5}$ )

this means you now have 20 x $\frac{5}{4}$

so you can simplify 20 and 4, this makes the 4 into a 1 and the 20 into a 5

you now have 5 x $\frac{5}{1}$

this is the same thing as 5 x 5

multiply it, and you can see that the answer is 25

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What is the area of this triangle?<br><br> A=bh2

Galina-37 [17]

Alright, so you know that the formula for the area of a triangle is a=bh/2

First start out by identifying your base and height.

Base is the side that the shape is "standing on", in this case, that would be 9cm.

Now, the height is the measurement that is perpendicular (90 degrees to the base), which in this case is 12 cm.

Now, plug these values into your formula

(12)(9) /2 = 54 cm squared! hope this helps :)

8 0

3 years ago

Can someone help me on problem number 8?

kramer

List the coordinates,
Coordinates are shown as (x,y)
If 2+2+7+7 is 16
You make those your side lengths
Coordinates are (9,7) (2,7) (2,5) (9,5)
they maybe slightly off because of the broken peice of paper

8 0

4 years ago

What is the value of m in the equation 1/2m - 3/4 = 16 when n=8

kow [346]

Answer:

answer is 33.5

Step-by-step explanation:

1/2m-3/4=16

1/2m=16+3/4

1/2m=64+3/4

1/2m=67/4

m=67/4×2

m=33.5

8 0

3 years ago

about 1300 organ specimens are preserved in fluid. Typically contains 7/10 alcohol and 3/10 water. if one jar contains 19 pins o

DerKrebs [107]

5.7 pints would be water

3/10 of 19 pints are water

19*(3/10) = 5.7

4 0

3 years ago

Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f

blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ .

b) Probability of finding the "odd" person in the $k$ th round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}$ .

c) Expected number of rounds: $\displaystyle \frac{2^{n - 1}}{n}$ .

Step-by-step explanation:

To decide the "odd" person, either of the following must happen:

There are $(n - 1)$ heads and $1$ tail, or
There are $1$ head and $(n - 1)$ tails.

Assume that the coins here all are all fair. In other words, each has a $50\,\%$ chance of landing on the head and a

The binomial distribution can model the outcome of $n$ coin-tosses. The chance of getting $x$ heads out of

The chance of getting $(n - 1)$ heads (and consequently, $1$ tail) would be $\displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n$ .
The chance of getting $1$ heads (and consequently, $(n - 1)$ tails) would be $\displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n$ .

These two events are mutually-exclusive. $\displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

Since the coins here are all fair, the chance of determining the "odd" person would be $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ in all rounds.

When the chance $p$ of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the $k$ th round: $(1 - p)^{k - 1} \cdot p$ . That's the same as the probability of getting one success after $(k - 1)$ unsuccessful attempts.

In this case, $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ . Therefore, the probability of succeeding on round $k$ round would be

$\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}$ .

Let $p$ is the chance of success on each round in a geometric distribution. The expected value of that distribution would be $\displaystyle \frac{1}{p}$ .

In this case, since $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ , the expected value would be $\displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}$ .

7 0

3 years ago