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2 years ago

In the recursive formula an = an−1 + 5, the value of the previous term is represented by

Mathematics

1 answer:

den301095 [7]2 years ago

6 0

Step-by-step explanation:

In the recursive formula an = an−1 + 5

$a_n = a_{n-1}+5$

Suppose if a_5 is the nth term then previous term is a_4

To get the previous term we need to subtract 1 from the n

So a_n is the nth term

$a_{n-1}$ is the previous term

Here 5 represents the common difference between two terms

So value of previous term is represented by

$a_{n-1}$

Answer : $a_{n-1}$

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Anna picked 200 apples at a orchard. 40% where green. How many green apples for annabelle pick.

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Step-by-step explanation:

I used a calculator to solve

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3 years ago

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing

kotegsom [21]

Answer:

D is correct

Step-by-step explanation:

Here, we want to select which of the options is correct.

The correct option is the option D

Since the die is unfair, we expect that the probability of each of the numbers turning up

will not be equal.

However, we should also expect that if we add the chances of all the numbers occurring together, then the total probability should be equal to 1. But this does not work in this case;

In this case, adding all the probabilities together, we have;

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3 years ago

Jessie read 1/3 of a book in the morning, and she read some more at night. By the end of the day, she still had 1/5 of the book

Mrac [35]

7/15 She read 7/15 at night

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3 years ago

Read 2 more answers

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne

Dvinal [7]

Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

$H_0: \bar x =12.7\\H_a: \bar x >12,7$

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = $\frac{6}{\sqrt{1000} } \\=0.1897$

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

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3 years ago

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

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1 year ago