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4 years ago

Write an expression for the nth term of... 2,1,0,-1...

Mathematics

1 answer:

Drupady [299]4 years ago

7 0

Answer:

$a_{n}$ = 3 - n

Step-by-step explanation:

There is a common difference between consecutive terms in the sequence, that is

1 - 2 = 0 - 1 = - 1 - 0 = - 1

This indicates the sequence is arithmetic with n th term

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 2 and d = - 1 , thus

$a_{n}$ = 2 - 1(n - 1) = 2 - n + 1 = 3 - n

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4 years ago

A recycling plant recycled 4.9 kg of paper and cardboard the first hour and 7.8 kg/h

Daniel [21]

Answer: R = {0kg, 59.5kg}

Step-by-step explanation:

When we have a function f(x) = y.

The set of the possible values of x is called the domain of the function, and the set of the possible values of y is called the range of the function,

In this case, we have that:

the plant recycled 4.9kg the first hour, and 7.8 kg per hour after.

So this function can be written as piecewise function, with one part for the first hour, and another part for the time after the first hour.

F(x) = 4.9kg*x if 0h < x ≤ 1h

F(x) = 4.9kg + 7.8kg/h*(x -1) if x ≥ 1h

We use (x - 1) for the second part because if x = 1 h, we must have the same result in both parts of the function.

Where x is the number of hours.

And we know that the domain is (0hours, 8 hours)

Then the minimum value of y will be:

F(0h) = 4.9kg*0h = 0kg.

And the maximum value in the range will be:

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Then the range is:

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7 0

3 years ago

A student randomly guesses on 10 true/false questions. use the binomial model to determine the probability that the student gets

lana [24]

Answer:

$P(5)=\frac{63}{256}$

Step-by-step explanation:

we are given

A student randomly guesses on 10 true/false questions

so,

$n=10$

there only true and false

so, probability getting true question is

$p=\frac{1}{2}$

so, probability getting false question is

$q=\frac{1}{2}$

the student gets 5 out of 10 questions right

so,

r=5

we can use binomial probability formula

$P(r)=\frac{n!}{r!(n-r)!} p^rq^{n-r}$

now, we can plug values

$P(5)=\frac{10!}{5!(10-5)!} (\frac{1}{2})^5(\frac{1}{2})^{10-5}$

we can simplify it

and we get

$P(5)=\frac{63}{256}$

4 0

3 years ago