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2 years ago

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In the partially shown sequence: …, 17711, A, B, 75025, …, each new term is the sum of the two previous terms. Find the whole nu

mber value of A. I GIVE BRAINLIEST!

1 answer:

yanalaym [24]2 years ago

4 0

Answer:

24513

Step-by-step explanation:

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Answer:

i'm pretty sure it is C..

Step-by-step explanation:

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2 years ago

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2 years ago

Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer

almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

$\to \mu=\frac{\sum xi}{n}$

$=22.8$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187$

In point b:

$\to \mu=\frac{\sum xi}{n}$

$=20.6875$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251$

In point c:

$\to \mu=\frac{\sum xi}{n}$

$=21$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179$

In point d:

$\to \mu=\frac{\sum xi}{n}$

$=20.8375$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\ =\sqrt{0.553166667} \\\\ =0.7438$

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Answer:

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