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3 years ago

What is the Volume of this Prism?

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

Answer:

240

Step-by-step explanation:

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Select all expressions that are equivalent to 6(b + 2) – b.

svlad2 [7]

Expressions equivalent would be 6b + 8 - b.

7 0

3 years ago

Factor.<br><br> 18b−32 <br><br> Enter your answer in the boxes.<br><br> [ ]([ ])

Mnenie [13.5K]

Here, GCF of 18 & 32 is 2, so take it as common factor,
18b - 32 = 2(9b - 16)

In short, Your Answer would be: 2(9b - 16)

Hope this helps!

3 0

3 years ago

Read 2 more answers

WILL MARK BRAINLEST, NEED THE ANSWER ASAP!!!!

PIT_PIT [208]

A. I’m so sorry I cant tell you how to do this part, get a formula chart from the Internet and just plug in the answers.
B. The slope of the ladder is also how it is slanted, if I’m reading it correctly, it’s there slanted part.
C. Much harder omgosh, imagine leaning a ladder up against a strait wall, and then pitting your weight I=on it backwards, here’s a hint you would fall flat on your back.

4 0

3 years ago

Solve for x in the equation x squared 20 x 100 = 36. x = –16 or x = –4 x = –10 x = –8 x = 4 or x = 16

spin [16.1K]

Possible values of x is -4 or -16

<h3>How to solve?</h3>

${x}^2$ + 20x +100 = 36

$\\{x}^2$ + 20x + 100-36 = 0

$\\{x}^2$ + 20x + 64 = 0

Using middle term split

In this method splitting of middle term in to two factors, In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term.

${x}^2$ + 4x + 16x + 64 = 0

x(x+4) + 16(x+4) = 0

taking common (x+4)

(x+4)(x+16) = 0

Either

x+4 = 0

⇒ x = -4

or (x+16) = 0

⇒ x = -16

Thus possible values of x is -4 or -16

To learn more about solving Quadratic Equations visit:

brainly.com/question/1279217

#SPJ4

7 0

2 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