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monitta
3 years ago
12

Can the probability of any event be 4? Yes or no explain your answer: i need help

Mathematics
1 answer:
Alika [10]3 years ago
7 0

Answer:

No.

Step-by-step explanation:

The maximum probability of something happening is 1, if there is no chance of another outcome.

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There are no specified Numbers provided

Step-by-step explanation:

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Liam is very excitrd to be getting a great deal on an iPad. it was originally $500 but he will only be paying $325. What Percent
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Given: an=3+an−1 and a1=5<br><br> What is the explicit rule for the arithmetic sequence?
Flura [38]

The explicit rule for the arithmetic sequence for the sequence is a(n) = 3n + 2.

<h3>What is a sequence?</h3>

It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.

We have given:

\rm a_n=3+a_n_-_1 and

\rm a_1=5

\rm a_n-a_n_-_1= 3

The above expression represents the common ratio:

d = 3

First term:

a = 5

The explicit rule for the arithmetic sequence:

a(n) = 5 + (n - 1)3

a(n) = 3n + 2

Thus, the explicit rule for the arithmetic sequence for the sequence is a(n) = 3n + 2.

Learn more about the sequence here:

brainly.com/question/21961097

#SPJ1

3 0
1 year ago
Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1
AysviL [449]

Answer:

a) The system has a unique solution for k\neq 6 and any value of h, and we say the system is consisted

b) The system has infinite solutions for k=6 and h=8

c) The system has no solution for k=6 and h\neq 8

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms (h), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]

The we can use the transformation r_0\rightarrow r_0 -2r_1, obtaining:

\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right].

Now we can start the analysis:

  • If k\neq 6 then, the system has a unique solution for any value of k, meaning that the last row will transform back to the equation as:

(k-6)x_2=h-8\\x_2=h-8/(k-6)

from where we can see that only in the case of k=6 the value of x_2 can not be determined.

  • if k=6 and h=8 the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

(k-6)x_2=h-8\\0=0 which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get x_1 as a function of x_2 o viceversa. Thus,  x_2 (x_1) is called a parameter since there are no constraints on what values they can take on.

if k=6 and h\neq 8 the system has no solution. Again by substituting in the equation resulting from the last row:

(k-6)x_2=h-8\\0=h-8 which is false for all values of h\neq 8 and since we have something that is not possible (0\neq h-8,\ \forall \ h\neq 8) the system has no solution

6 0
3 years ago
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