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artcher [175]
2 years ago
13

A business has $25,000 to spend on training sessions for its employees. It wants 45 of its employees to attend. The business wan

ts to send as many employees as it can to a technology training. The technology training costs $1,000 per person. The customer service training costs $500 per person. Create a system of equations that models how many of each type of training the business should purchase. 1,000x + 500y = 45 x + y = 25,000 1,000x + 500y = 25,000 x + y = 45 1,000x + y = 45 x + 500y = 25,000 x + 500y = 45 1,000x + y = 25,000
Mathematics
1 answer:
lilavasa [31]2 years ago
5 0

Answer: x+y=45\\\\1000 x + 500y = $2500

Step-by-step explanation:

Let x =  Number of employees taking technology training

y= Number of employees taking customer service training

Given, The technology training costs $1,000 per person. The customer service training costs $500 per person.

Total cost = 1000 x + 500y

Since, Total cost = $25,000 and total employee to attend training= 45 .

That means , the required equations are:

x+y=45\\\\1000 x + 500y = $2500

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The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{
Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

 

km

3

kg

​

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

 

km

3

⋅1.03⋅10

12

 

km

3

kg

​

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

​

​

 

=

6⋅10

24

kg

1.3802⋅10

21

kg

​

=

6⋅10

3

1.3802

​

=

6000

1.3802

​

=0.0002300

3

​

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0
1 year ago
Nswer two questions about Systems A AA and B BB: System A AA start text, end text System B BB { − 3 x + 12 y = 15 7 x − 10 y = −
rjkz [21]

Answer:

(Choice C) C Replace one equation with a multiple of itself

Step-by-step explanation:

Since system A has the equations

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system B has the equations

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To get system B from system A, we notice that equation -x + 4y = 5 is a multiple of -3x + 12y = 15 ⇒ 3(-x + 4y = 5) = (-3x + 12y = 15).

So, (-x + 4y = 5) = (1/3) × (-3x + 12y = 15)

So, we replace the first equation in system B by 1/3 the first equation in system A to obtain the first equation in system B.

So, choice C is the answer.

We replace one equation with a multiple of itself.

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