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3 years ago

Help pls

Mathematics

2 answers:

Montano1993 [528]3 years ago

6 0

Answer:

The partial sums are increasing in value.

The series converges to the same value as what the partial sums approach.

The partial sums appear to approach 0.33.

The series might converge to One-third.

Step-by-step explanation:

mariarad [96]3 years ago

4 0

Answer:

The partial sums are increasing in value.

The series converges to the same value as what the partial sums approach.

The partial sums appear to approach 0.33.

The series might converge to One-third.

Step-by-step explanation:

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Find the value of S for an infinite geometric series with r= 1/6 and a1=10

Lelu [443]

S=a1/1-r=10 x 6/5 = 12

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3 years ago

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following p

aev [14]

Answer:

P(X < 3) = 0.7443

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success. Also, given n = 6, p = 0.3.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6

r = number of success = less than 3

p = probability of success which in our question is 0.3.

LET X = a random variable

So, it means X ~ $Binom(n=6, p=0.3)$

Now, Probability that X is less than 3 = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{6}{0}0.3^{0} (1-0.3)^{6-0}+ \binom{6}{1}0.3^{1} (1-0.3)^{6-1}+ \binom{6}{2}0.3^{2} (1-0.3)^{6-2}$

= $1 \times 1 \times 0.7^{6} +6 \times 0.3^{1} \times 0.7^{5} +15 \times 0.3^{2} \times 0.7^{4}$

= 0.11765 + 0.30253 + 0.32414 = 0.7443

Therefore, P(X < 3) = 0.7443.

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4 years ago

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Answer:

The table doesn't display a linear relationship because the slope doesn't stay constant throughout the whole table.

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web based software company is interested in estimating the proportion of individuals who use the Firefox browser. In a sample of

nata0808 [166]

<h2>Answer with explanation:</h2>

Let $\hat{p}$ denotes the sample proportion.

As per given , we have

n= 200

$\hat{p}=\dfrac{30}{200}=0.15$

Critical value for 99% confidence : $z_{\alpha/2}=2.576$

Confidence interval for population proportion :-

$\hat{p}\pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.15\pm (2.576)(\sqrt{\dfrac{0.15(1-0.15)}{200}})\\\\\approx0.15\pm0.065\\\\=(0.15-0.065,\ 0.15+0.065)\\\\=(0.085,\ 0.215)$

Hence, a 99% confidence interval for the proportion of all individuals that use Firefox: $(0.085,\ 0.215)$

The lower limit on the 99% confidence interval = 0.085

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Alla [95]

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Step-by-step explanation:

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