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3 years ago

ONLY GIVING BRAINLYEST IF YOU ANSWER QUICKLY AND CORRECTLY

Mathematics

2 answers:

ryzh [129]3 years ago

7 0

Answer:

A: 7 was not multiplyed by Y

Step-by-step explanation:

I learned it in math class last year.

aalyn [17]3 years ago

6 0

Answer:

option A is correct answer

Step-by-step explanation:

=> 7 was not multiplied by y

=> hope it helps

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Help me pleaseeeee.

Natali [406]

Answer:

7.) Line, point

8.) Line

9.) Line, point, rotational

10.) Rotational

11.) None

12.) Line

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

Find a solution to the initial value problem,<br> y″+12x=0, y(0)=2,y′(0)=−1.

levacccp [35]

Answer:

y = -2*x^3 - x + 2

Step-by-step explanation:

We want to solve the differential equation:

y'' + 12*x = 0

such that:

y(0) = 2

y'(0) = -1

We can rewrite our equation to:

y'' = -12x

if we integrate at both sides, we get:

$\int {y''} \, dx = y'= \int {-12x} \, dx$

Solving that integral we can find the value of y', so we will get:

y' = -12* (1/2)*x^2 + C = -6*x^2 + C

where C is the constant of integration.

Evaluating y' in x = 0 we get:

y'(0) = -6*0^2 + C = C

and for the initial value problem, we know that:

y'(0) = -1

then:

y'(0) = -1 = C

C = -1

So we have the equation:

y' = -6*x^2 - 1

Now we can integrate again, to get:

y = -6*(1/3)*x^3 - 1*x + K

y = -2*x^3 - x + K

Where K is the constant of integration.

Evaluating or function in x = 0 we get:

y(0) = -2*0^3 - 0 + K

y(0) = K

And by the initial value, we know that: y(0) = 2

Then:

y(0) = 2 = K

K = 2

The function is:

y = -2*x^3 - x + 2

4 0

3 years ago

Simplify the following 7 times the square root of 54 - 2 times the square root of 24

EleoNora [17]

The first step for solving this problem is to simplify the radical
7 $\sqrt{54 - 2}$ x 2√6
Calculate the product
14 $\sqrt{(54 - 2) X 6}$
Subtract the numbers
14 $\sqrt{56 x 6}$
Simplify the radical
14 x 2 $\sqrt{13 x 6}$
Multiply the numbers
14 x 2√78
Finally,, calculate the product for your final answer
28√78
Let me know if you have any further questions
:)

4 0

4 years ago

Solve for y: 5( 3y+4) = 6( 2y-2/3)

N76 [4]

Answer: y= -8

Step-by-step explanation:

5( 3y+4) = 6( 2y-2/3)
15y+20=12y-4
15y-12y=-4-20
3y=-24
y=-24/3
y= -8

4 0

2 years ago