5X/2 = 20<br> WHAT IS THE ANSWER TO X

Set A of six numbers has a standard deviation of 3 and set B of four numbers has a standard deviation of 5. Both sets of numbers

Alenkinab [10]

Given:

$\sigma_A=3$

$n_A=6$

$\sigma_B=5$

$n_B=4$

$\overline{x}_A=\overline{x}_B$

To find:

The variance. of combined set.

Solution:

Formula for variance is

$\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}$ ...(i)

Using (i), we get

$\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}$

$(3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$54=\sum (x_i-\overline{x}_A)^2$

Similarly,

$\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}$

$(5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$100=\sum (x_i-\overline{x}_B)^2$

Now, after combining both sets, we get

$\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}$

$\sigma^2=\dfrac{54+100}{6+4}$

$\sigma^2=\dfrac{154}{10}$

$\sigma^2=15.4$

Therefore, the variance of combined set is 15.4.

7 0

3 years ago

Company A has 400 employees and Company B has 500 employees. Among these employees, there are 50 married couples, each consistin

abruzzese [7]

Answer:0,0025

Step-by-step explanation:

P(select a married couple) = P(person selected from Company A is one of the 50 with spouse at Company B) x P(person selected from Company B spouse of person already selected from Company A)

Now you can substitute the respective probabilities .

= (50/400) x (1/500)

= 0,0025

8 0

3 years ago

a. Show that the following statement forms are all logically equivalent. p → q ∨ r, p ∧ ∼q → r, and p ∧ ∼r → q b. Use the logica

slava [35]

Answer:

(a) if n is prime, then n is odd or n is 2

(b) if n is prime and n is not odd, then n is 2

(c) if n is prime and n is not 2, then n is odd

Step-by-step explanation:

a) p → q ∨ r

b) p ∧ ∼q → r

c) p ∧ ∼r → q

Lets show that (a) implies (b) and (c). (a) says that if property p is true, then either q or r is true, thus, if p is true we have:

If the condition of (b) applies (thus q is not true), we need r to be true because either q or r were true because we are assuming (a) and p. Hence (b) is true
If the condition of (c) applies (r is not true), since either r or q were true due to what (a) says, then q neccesarily is true, hence (c) is also true.

Now, lets prove that (b) implies (a)

If p is true and property (b) is true, then if q is true, then either q or r are true thus (a) is correct. If q is not true, then property (b) claims that, since p is true and q not, r has to be true, therefore (a) is valid in this case as well, hence (a) is also true.

(c) implies (a) can be proven with similar argument, changing (b) for (c), q for r and r for q.

With this we prove that the 3 properties are equivalent.

For the rest of the exercise, we have

property p: n is prime
property q: n is odd
property r: n is 2

Translating this, we obtain (a), (b) and (c)

(a) if n is prime, then n is odd or n is 2

(b) if n is prime and n is not odd, then n is 2

(c) if n is prime and n is not 2, then n is odd

7 0

4 years ago

You must select a team from a group of 6 boys and 10 girls. How many ways can you select 2 boys and 5 girls?

Keith_Richards [23]

Using the combination formula, it is found that you can select 2 boys and 5 girls in 2,520 ways.

In this problem, Joao and Elisa would be the same team as Elisa and Joao, hence the order is not important and the <em>combination formula</em> is used to solve this question.

<h3>What is the combination formula?</h3>

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by: