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3 years ago

How do i write this proof?

Mathematics

1 answer:

storchak [24]3 years ago

6 0

It is a rhombus because:
• The figure has four sides
• The acute angles are equal and opposite each other
• The obtuse angles are equal and opposite each other
• The figure is a parallelogram
Hope this helps!

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Solve the simultaneous equation 5x+3y=23 2x+4y=12

uranmaximum [27]

() = Fraction

Answer:

1. x= −(3 /5 y)+ (23 /5)

2. x=−2y+6

Step-by-step explanation:

Let's solve for x.

5x+3y=23

Step 1: Add -3y to both sides.

5x+3y+−3y=23+−3y

5x=−3y+23

Step 2: Divide both sides by 5.

5x /5 = −((3y+23) /5)

x= −(3 /5 y)+ (23 /5)

------------------------------------------------

Let's solve for x.

2x+4y=12

Step 1: Add -4y to both sides.

2x+4y+−4y=12+−4y

2x=−4y+12

Step 2: Divide both sides by 2.

(2x /2 )= −((4y+12) /2)

x=−2y+6

Hope it helped!

5 0

3 years ago

In a taste test, each sample serving of a new drink is about 50 milliliters The drink comes in a 1 liter

expeople1 [14]

Answer:

Step-by-step explanation:

it is 25 $25+125=50$

7 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

A bicycle’s speedometer measures speed to the nearest 0.1 miles per hour. Which is the most appropriate way to report speed usin

Rina8888 [55]

The bicycle`s speedometer measures the speed to the nearest 0.1 mph. So 17.15 mph is not appropriate way to report speed, because then it should measure speed to the nearest 0.05 mph. Also 17 mph and 20 mph are whole numbers and it does not have to be so accurate.
Answer: B. 17.1 mph

5 0

4 years ago

Please help i have been trying to figure this out for like an hour

DiKsa [7]

Answer:

que rayosllpplpppewwe

7 0

3 years ago