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3 years ago

Help pls!

Mathematics

1 answer:

ohaa [14]3 years ago

6 0

JKL is 54 cm.

JL is 24 cm.

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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

4 years ago

A fair coin is tossed 5000 times. What can you say about getting the outcome of exactly 2500 tails

WINSTONCH [101]

Step-by-step explanation:

You can't expect to get exactly 2500 out of 5000 tosses more than a few times . You will come pretty close, but that's only good in horseshoes.

Of course I'm answering this on the basis of a computer language and not actually performinig this a million tmes, each part of a million consisting of 5000 tosses.

Simulations and not completely unbiased, but based on experience, 5000 is a very small number and getting 2500 more than a couple of times is unlikely

6 0

3 years ago

Read 2 more answers

A student takes an exam containing 18 multiple choice questions. The probability of choosing a correct answer by knowledgeable g

lawyer [7]

When you have to repeatedly take the same test, with constant probability of succeeding/failing, you have to use Bernoulli's distribution. It states that, if you take $n$ tests with "succeeding" probability $p$ , and you want to "succeed" k of those n times, the probability is

$\displaystyle P(n,k,p) = \binom{n}{k}p^k(1-p)^{n-k}$

In your case, you have n=18 (the number of tests), and p=0.3 (the probability of succeeding). We want to succeed between 8 and 12 times, which means choosing k=8,9,10,11, or 12. For example, the probability of succeeding 8 times is

$\displaystyle P(18,8,0.3) = \binom{18}{8}(0.3)^8(0.7)^{10}$