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tekilochka [14]
3 years ago
10

A triangle is formed using three given side lengths. Do these lengths always forma unique triangle? Explain.​

Mathematics
1 answer:
ollegr [7]3 years ago
6 0
No, these do not always forma unique triangles because it depends on what size your sides are because if you have 3 sides of the same length then that’s a equilateral and if you have two sides of the same length and one side that’s not then that would be an Isosceles triangle and if you had no sides of the same length then that would be a scalene triangle.
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lbvjy [14]
.0001*5.5 + 1.5*x = 2
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3 years ago
Write the first 4 terms of the arithmetic sequence With first term 37 and common difference 9
alexandr402 [8]

Answers:

first term = 37

second term = 46

third term = 55

fourth term = 64

==========================================

Explanation:

Start at 37. Add 9 to this term to get the second term. Add 9 to that result to get the next term, and so on.

first term = 37

second term = 37+9 = 46

third term = 46+9 = 55

fourth term = 55+9 = 64

-----------

The nth term formula is

a(n) = 9n+28

It can be found by plugging a = 37 and d = 9 into the general form

a(n) = a + d*(n-1)

4 0
3 years ago
Read 2 more answers
Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}


In which

x is the number of sucesses


e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164


0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671

0.671 = 67.1% probability that neither contains a missing pulse

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Answer: The median of the date you gave would be -71. Hope that helps!

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The quotient of c and 6
Elena L [17]

Answer:

"a quotient of a number and 6" refers to n6 or n÷6 .

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