All categories

2 years ago

Find the first six terms of the sequence. a1 = -1, an = 2 • an-1

Mathematics

2 answers:

kompoz [17]2 years ago

7 0

Answer:

The first six terms of the sequence are: -1, -2, -4, -8, -16 and -32

Step-by-step explanation:

The sequence is:

$a_1=-1\\a_n=2*a_{n-1}$

So, for finding the first six terms of the sequence, we need to replace n by the numbers 1 to 6 and we get:

$a_1=-1\\a_2=2*a_{2-1} =2*a_1\\a_3=2*a_{3-1} =2*a_2\\a_4=2*a_{4-1} =2*a_3\\a_5=2*a_{5-1} =2*a_4\\a_6=2*a_{6-1} =2*a_5\\$

That means that the term n is 2 times the last term. Then, replacing the values of $a_1, a_2, a_3, a_4$ and $a_5$ , we get:

$a_1=-1\\a_2=2*(-1)=-2\\a_3=2*(-2)=-4\\a_4=2*(-4)=-8\\a_5=2*(-8)=-16\\a_6=2*(-16)=-32\\$

So, The first six terms of the sequence are: -1, -2, -4, -8, -16 and -32

allochka39001 [22]2 years ago

3 0

The prompt says that the value at n is equal to 2 times the previous value.
-1, -2, -4, -8, -16, -32

You might be interested in

I need help right now please

Hoochie [10]

8/1 + (x-1)/(x+4)
You need the same common denominator (bottom number) to be able to add together so multiply 8 by (x+4) to both bottom and top

8 (x+4)/(x+4) + (x-1)/(x+4)
Add
8 (x+4)+(x-1)/ (x+4)
Distribute the 8 to (x+4)
8x+32+x-1 / x+4
Simplify

9x+31/x+4
:)

8 0

2 years ago

There are 64 cookies in a jar. The probability of randomly choosing oatmeal cookies from the jar is 37.5%. How many of the cooki

Rufina [12.5K]

Answer:

the number of cookies that are oatmeal cookies is 24

Step-by-step explanation:

Given that

There are 64 cookies in a jar

And, the probability of randomly selecting the oatmeal cookies is 37.5%

We need to find out the number of cookies that are oatmeal cookies

So, the number of cookies that are oatmeal cookies are

= 64 cookies × 37.50%

= 24 cookies

hence, the number of cookies that are oatmeal cookies is 24

6 0

2 years ago

Assume that when adults with smartphones are randomly selected, 58% use them in meetings or classes. If 10 adult smartphone us

Tasya [4]

Answer:

79.85% probability that at least 5 of them use their smartphones in meetings or classes.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they use their smartphone during meetings or classes, or they do not. The probability of an adult using their smartphone in these situations are independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability 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.

Assume that when adults with smartphones are randomly selected, 58% use them in meetings or classes.

This means that $p = 0.58$

10 adults selected.

This means that $n = 10$

Find the probability that at least 5 of them use their smartphones in meetings or classes.

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 5) = C_{10,5}.(0.58)^{5}.(0.42)^{5} = 0.2162$

$P(X = 6) = C_{10,6}.(0.58)^{6}.(0.42)^{4} = 0.2488$

$P(X = 7) = C_{10,7}.(0.58)^{7}.(0.42)^{3} = 0.1963$

$P(X = 8) = C_{10,8}.(0.58)^{8}.(0.42)^{2} = 0.1017$

$P(X = 9) = C_{10,9}.(0.58)^{9}.(0.42)^{1} = 0.0312$

$P(X = 10) = C_{10,10}.(0.58)^{10}.(0.42)^{0} = 0.0043$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2162 + 0.2488 + 0.1963 + 0.1017 + 0.0312 + 0.0043 = 0.7985$

79.85% probability that at least 5 of them use their smartphones in meetings or classes.

3 0

3 years ago

Which set of integers is not a Pythagorean triple?

In-s [12.5K]

<span> B) 14, 46, 48 is not a Pythagorean triplet.</span>

7 0

3 years ago

Which of the following describes the domain of y = tan x, where n is any integer? A.x≠2nπ B. x≠π/2+nπ C. x≠nπ D.x≠nπ/2

lidiya [134]

$\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}$

therefore, that fraction will become undefined if the denominator ever turns to 0.

now, recall that cos(π/2) is 0, and cos(3π/2) is 0, and cos(5π/2) is also zero and so on, thus

$\bf tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{cos\left( \frac{\pi }{2} \right)}\implies tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{0}
\\\\\\
tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{cos\left( \frac{3\pi }{2} \right)}\implies tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{0}
$

$\bf tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{cos\left( \frac{5\pi }{2} \right)}\implies tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{0}
\\\\\\
tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{cos\left( \frac{7\pi }{2} \right)}\implies tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{0}$

$\bf tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{cos\left( \frac{\pi }{2}+\pi \right)}\implies tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{0}
\\\\\\
domain\implies \{x|x\in \mathbb{R}, ~~x\ne \frac{\pi }{2}+n\pi,~~n\in \mathbb{Z} \}$

8 0

3 years ago