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3 years ago

Use the PACED decision-making process to _____.

1 answer:

6 0

Use the PACED decision-making process to make a decision or make a choice. The steps to follow when making a decision can be easily remembered when you use the word PACED:
P - What is the PROBLEM - Why is there a need to make a choice?
A - What are the ALTERNATIVES - What are the possible choices available?
C - Establish the CRITERIA - Why is one choice better than the other?
E - EVALUATE the alternatives - How well does each choice meet the criteria?
D - Make a DECISION - What is the best choice?

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Jessica has two strings. the first string measures 80 cm and the second string measures 64cm. she wants to cut the strings into

omeli [17]

The highest common denominator between 80 and 64 is 16. This means no string can be cut longer than 16 inches for them all to have the same length.

5 0

3 years ago

A. Solve the differential equation <img src="https://tex.z-dn.net/?f=y%27%3D2x%20%5Csqrt%7B1-y%5E2%7D%20" id="TexFormula1" title

kirill [66]

$y' = \frac{dy}{dx}$

seperable differential equations will have the form
$\frac{dy}{dx} = F(x) G(y)$

what you do from here is isolate all the y terms on one side and all the X terms on the other
$\frac{dy}{G(y)} = F(x) dx$
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
$\int \frac{1}{G(y)} dy = \int F(x) dx

$

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it

In your problem,
$G(y) = \sqrt{1-y^2}

F(x) = 2x$

so all you need to integrate is
$\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx$

5 0

3 years ago

George bought socks for "$"300 and planned to sell them for $6 a pair. write an inequality to find the minimum number of socks h

Anit [1.1K]

The inequality 6s - 300 > 210 represents the minimum number of socks (s) when George bought socks for "$"300 and planned to sell them for $6 a pair, and want to make a profit of more than 210.

In the question, we are given that George bought socks for $300.

Therefore, George's total cost = $300.

We assume the number of pairs of socks George sells to be s.

The unit price per pair quoted by George = $6

Therefore, George's total sales = $6s.

We know that the profit over a transaction is given as the difference between sales and costs, that is,

The profit = The total sales - the total costs,

or, profit = 6s - 300.

In the question, we are required to make a profit of more than 210.

This can be shown by the inequality,

6s - 300 > 210.

Thus, the inequality 6s - 300 > 210 represents the minimum number of socks (s) when George bought socks for "$"300 and planned to sell them for $6 a pair, and want to make a profit of more than 210.

Learn more about inequalities at

brainly.com/question/24372553

#SPJ4

5 0

1 year ago

Convert the following common fractions to decimals. Round to three decimal places if necessary. 5/6 = A)0.83 B)5.6 C)0.56 D)0.83

Umnica [9.8K]

Answer:

Step-by-step explanation:

If we divide 5 by 6 we'll come up with 8.333 repeating. It is necessary that we round to three decimal places since we have an endless amount of digits that can only be represented with 3 digits

7 0

3 years ago

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

4 0

2 years ago