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3 years ago

What is 11.35 divided by 0.35?

Mathematics

2 answers:

sashaice [31]3 years ago

6 0

Answer:

32.4285714

Step-by-step explanation:

11.35/0.35=32.4285714

This is the answer however if your question has asked to round it to two decimal places the answer would be 32.43

i hope this helps

Fantom [35]3 years ago

3 0

Answer:

32.43 (rounded)

Step-by-step explanation:

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nika2105 [10]

Answer:

I think that answer is D option

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3 years ago

A. How many ways can you distribute $4$ different balls among $4$ different boxes? B. How many ways can you distribute $4$ ident

zysi [14]

Answer:

The answer is explained below

Step-by-step explanation:

A. How many ways can you distribute 4 different balls among 4 different boxes

The number of ways in which 4 different balls can be distributed among 4 different boxes is $4^4=256\ ways$

B. How many ways can you distribute 4 identical balls among 4 identical boxes?

The number of ways in which 4 identical balls can be distributed among 4 identical boxes = P(4,1) + P(4,2) +P(4,3) + P(4,4) = 1 + 2 + 1 + 1 = 5 ways

Where P(k,n) is the number of partitions that k can be divided into n parts

P(4,1) = 4 = 1

P(4,2) = 1 + 3, 2+2 = 2

P(4,3) = 1 + 1 + 2 = 1

P(4,4) = 1 + 1 + 1 + 1 = 1

C. How many ways can you distribute 4 identical balls among 4 different boxes

The number of ways in which 4 identical balls can be distributed among 4 different boxes = $C(4\ ball+4\ box-1,4\ box-1)=C(7,3)=\frac{7!}{(7-3)!3!}=35\ ways$

7 0

3 years ago

4e^2x=5 <br><br> X approximately = ?

olga nikolaevna [1]

Answer:

5 = 5

Explanation:

5 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

Graph these 2 equations for me?? y=5(1/2)^x +4 y=4(1/6)^(x+2)

olchik [2.2K]

The graphs of the two equation are attache. The red curve represents the graph of the equation $y=5(1/2)^x +4$ while the blue curve represents the graph of the equation $y=4(1/6)^{(x+2)}$ .

8 0

4 years ago

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