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BARSIC [14]
3 years ago
12

Pls help:( i’ll mark brainlest!

Mathematics
1 answer:
12345 [234]3 years ago
8 0

;-;

Step-by-step explanation:

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Hich decimal represents the shaded portion?
svlad2 [7]
.3, 3 out of 10 are covered making .3, 30%
8 0
3 years ago
a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

|s_n -s| < \epsilon for all n, and that implies that the series converges to a value, s.

And this complete the proof.

<em>Part c</em>

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where s_n represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most b_{n+1} by the following theorem:

<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any  two consecutive partial sums sn and sn+1.

\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

And this complete the proof.

5 0
3 years ago
What's 14 m = ____ mm?<br> 790 cm = ___ m?<br> 782 km = 782,000 ___?
MrRissso [65]
14 m: 14,000 mm
790 cm : 7.9 m
 782 km : 782,000 grams

8 0
3 years ago
The mean is 47.1 and the standard deviation is 9.5 for a population. Using the Central Limit Theorem, what is the standard devia
MaRussiya [10]

Answer:

The standard deviation of the distribution of sample means for samples of size 60 is of 1.2264.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Standard deviation is 9.5 for a population.

This means that \sigma = 9.5

Sample of 60:

This means that n = 60

What is the standard deviation of the distribution of sample means for samples of size 60?

s = \frac{\sigma}{\sqrt{n}} = \frac{9.5}{\sqrt{60}} = 1.2264

The standard deviation of the distribution of sample means for samples of size 60 is of 1.2264.

4 0
2 years ago
The dimensions of a carriage are:
jasenka [17]

Answer:

Step-by-step explanation:

length * width * height = 1050 cubic ft

(10x + 5)* (2x -1) * (x +3) = 1050

(10x + 5 )(2x-1) = 10x*(2x - 1) + 5*(2x -1)

                       = 10x*2x - 10x*1 + 5*2x  - 1*5

                       = 20x² - 10x + 10x - 5 = 20x²  - 5

(10x + 5)(2x -1)(x+3) = (20x² -5) *(x + 3)

                               =20x²*(x +3) - 5*(x +3)

                               = 20x³ + 60x² - 5x - 15

20x³ + 60x² - 5x - 15 = 1050

20x³ + 60x² - 5x - 15 - 1050 = 0

20x³ + 60x² - 5x - 1065 = 0

By trial and error method , (x  -3 ) is a factor

Synthetic method:

3     ( 20    60       -5           -1065

      (          60        360        1065

         20   120        355           0

20x² + 120x + 355 = 0

Divide the equation by 5

4x² + 24x + 71 = 0 . Solving this equation gives imaginary number. So, ignore these values.

So, value of x = 3

height = 3+ 3 = 6 ft

Length = 10*3 + 5 = 30 + 5 = 35ft

Width = 2x - 1 = 2*3 - 1 = 6 - 1 = 5 ft

Check: length * width * height = 35 * 5 * 6 = 1050 cubic ft

5 0
3 years ago
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