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Daniel [21]
3 years ago
5

The legs of an isosceles triangle measure ( 2 x^4 + 2 x − 1 ) units each. The perimeter of the triangle is ( 5 x^4 − 2 x^3 + x −

3 ) units. Write a polynomial (in simplest standard form) that represents the measure of the base of the triangle.
Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

The base is x^4 -2x^3 -3x-1

Step-by-step explanation:

We know the perimeter of a triangle is the sum of the three sides

P = s1+s2+s3

We know the perimeter is  5 x^4 − 2 x^3 + x − 3

and two of the legs are 2 x^4 + 2 x − 1  since it is an isosceles triangle

P = 2s1 + s3

Subtract 2s1 from each side

P-2s1 =2s1 +s3 -2S1

P -2s1 =s3

Substituting what we know

5 x^4 − 2 x^3 + x − 3 - 2(2 x^4 + 2 x − 1) = s3

Distribute the -2

5 x^4 − 2 x^3 + x − 3 - 4 x^4 -4 x + 2 = s3

Combine like terms

5 x^4-4x^4 − 2 x^3 + x  -4 x -3+ 2 = s3

x^4 -2x^3 -3x-1 =s3

The base is x^4 -2x^3 -3x-1

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What products result in a difference of squares?
lesya [120]

Answer:

cubes?

Step-by-step explanation:

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 kth 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:

<h3>a)</h3>

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.

<h3>b)</h3>

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 kth 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}.

<h3>c)</h3>

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
You guessed the height of the building to be 23 feet, but it was actually 35 feet. What was your percent error?
Juli2301 [7.4K]

Answer:

The error was  28.575%, rounded to 29%.

Step-by-step explanation:

(35-25/35)

4 0
3 years ago
Salma bought
Julli [10]
It should be p=15n
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3 0
3 years ago
Read 2 more answers
twelve students in a class average 70% on a certain test. eighteen others average 80%. what is the overall average of the thirty
tia_tia [17]
Given:
12 students average 70%
18 students average 80%

What is the overall average of the 30 students as a percent?

12 x 70% = 840%
18 x 80% = 1440%

840% + 1440% = 2280%

2280% / 30 = 76%

The overall average of the 30 students is 76%.
6 0
3 years ago
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