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3 years ago

Please help with this

Mathematics

1 answer:

tresset_1 [31]3 years ago

5 0

Answer:

1, 2, 3, 6, 9, 18, 27, 54

Step-by-step explanation:

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Companies use employee training for various reasons, including employee loyalty, certification, quality, and process improvement

Radda [10]

Answer:

0.3081 = 30.81% probability that the company uses training for employee retention and not for process improvement

Step-by-step explanation:

56% employee retention.

38% for process improvement. Of those, 89% for employee retention.

a. What is the probability that the company uses training for employee retention and not for process improvement

The 56% is composed by:

x of 100 - 38 = 72%(not for process improvement)

89% of 38%(for process improvement).

We want to find x. So

$0.72x + 0.89*0.38 = 0.56$

$0.72x = 0.56 - 0.89*0.38$

$x = \frac{0.56 - 0.89*0.38}{0.72}$

$x = 0.3081$

0.3081 = 30.81% probability that the company uses training for employee retention and not for process improvement

5 0

3 years ago

Is 2/3(3/4x-3/2) equal to 1/2x -1

ziro4ka [17]

Answer:

Step-by-step explanation:

that would be the anwer

6 0

3 years ago

Read 2 more answers

Plz help me plz fugugughfufgfhvjgu

Anon25 [30]

Answer:

B is the only one i know, sorry.

Step-by-step explanation:

Hope this helped

8 0

3 years ago

PLEASE HELLP Eric plays basketball and volleyball for a total of 95 minutes every day. He plays basketball for 25 minutes longer

Anna11 [10]

Answer:

Part A:

(1) x + y = 95

(2) x = y + 25

Part B:

The number of minutes Eric spends playing volleyball each day is 35 minutes

Part C:

It is not possible for Eric to have spent exactly 35 minutes playing basketball

Step-by-step explanation:

The total time Eric plays basketball and volleyball = 95 minutes

The time duration Eric plays basket ball = x

The time duration Eric plays volleyball = y

Part A:

The pair of relationships between the number of minutes Eric plays basketball (x) and the number of minutes he plays volleyball (y) are;

(1) x + y = 95

(2) x = y + 25

Part B:

By substituting the value of x in equation (2) into equation (1), we have;

x + y = (y + 25) + y = 95

2·y + 25 = 95

2·y = 95 - 25 = 70

y = 70/2 = 35 minutes

Therefore, Eric spends 35 minutes playing volleyball every day

Part C:

It is not possible for Eric to have spent only 35 minutes playing basketball because, given that he plays basketball for 25 minutes longer than he plays volley, the number of minutes he spends playing volleyball will then be given as follows;

x = y + 25

35 = y + 25

y = 35 - 25 = 10 minutes

The total time = x + y = 10 + 35 = 45 minutes ≠ 95 minutes.

3 0

3 years ago

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

8 0

2 years ago