I NEED HELP PLEASE HELP

Suppose that 7 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at ra

katen-ka-za [31]

Answer:

(a) 350

(b) 175

(c) 21

(d) 196

Step-by-step explanation:

Number of females = 7

Number of males = 5

Total ways of selecting r items from n items is

$^nC_r=\dfrac{n!}{r!(n-r)!}$

(a)

Total ways of selecting 3 females and 2 males.

$\text{Total ways}=^7C_3\times ^5C_2$

$\text{Total ways}=\dfrac{7!}{3!(7-3)!}\times \dfrac{5!}{2!(5-2)!}$

$\text{Total ways}=35\times 10$

$\text{Total ways}=350$

(b)

Total ways of selecting 4 females and 1 male.

$\text{Total ways}=^7C_4\times ^5C_1$

$\text{Total ways}=32\times 5$

$\text{Total ways}=175$

(c)

Total ways of selecting 5 females.

$\text{Total ways}=^7C_5\times ^5C_0$

$\text{Total ways}=21\times 1$

$\text{Total ways}=21$

(d)

Total ways of selecting at least 4 females.

Total ways = 4 females + 5 females

$\text{Total ways}=175+21$

$\text{Total ways}=196$

6 0

3 years ago

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

kramer

Answer:

(a)

$\bar x = 0.0075$

$s = 0.0049$

(b)

B. The sample is too small to make judgments about skewness or symmetry.

Step-by-step explanation:

Given:

$n = 8$

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 \ \end{array}$

Solving (a):

First, calculate the difference between the recorded TBBMC for both operators:

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 &{|1 - 2|} &0.003 & 0.015 & 0.006 & 0.004 & 0.002 & 0.010 & 0.005 & 0.015 \ \end{array}$

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.

The mean is then calculated as:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{0.003 + 0.015 + 0.006 + 0.004 + 0.002 + 0.010 + 0.005 + 0.015}{8}$

$\bar x = \frac{0.06}{8}$

$\bar x = 0.0075$

Next, calculate the standard deviation (s).

This is calculated using:

$s = \sqrt{\frac{\sum (x - \bar x)^2}{n}}$

So, we have

$s = \sqrt\frac{(0.003 - 0.0075)^2 + (0.015 - 0.0075)^2+ (0.006- 0.0075)^2+ ....... + (0.005- 0.0075)^2+ (0.015- 0.0075)^2 }{8}$ $s = \sqrt\frac{0.00019}{8}$