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Anton [14]
3 years ago
13

1. The number of employees at a certain company is 1440 and is increasing

Mathematics
1 answer:
adelina 88 [10]3 years ago
6 0

Answer:

y = 1440 e^{0.015 t}

y(9) = 1440 e^{0.015*9}= 1648.133

So then after 9 years we will have approximately 1649 number of employees

Step-by-step explanation:

For this case we want to model the number of employees and we need to use an exponential model given by this general expression:

y = y_o e^{rt}

For this case the initial amount is y_o = 1440 and the rate r =0.015

And then the model would be given by

y = 1440 e^{0.015 t}

And if we find the value for t =9 years we got:

y(9) = 1440 e^{0.015*9}= 1648.133

So then after 9 years we will have approximately 1649 number of employees

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Reika [66]

Answer:

x=40

Step-by-step explanation:

2x+7+x-10+63=180

3x=120

x=40

7 0
3 years ago
Mr. Mole's burrow lies 5 55 meters below the ground. He started digging his way deeper into the ground, descending 3 33 meters e
Alexeev081 [22]

Answer:

184,815

Step-by-step explanation:

so my answer my not be right so i would say double check but anyway just multiply 3333 and 555 and this should be your answer.

7 0
3 years ago
What simple interest rate would allow $6000 to grow to an amount of $14550 in 10 years?
Harman [31]

Answer:

\boxed{ \bold{ \huge{  \boxed{ \sf{ \: 14.25 \: \% \: }}}}}

Step-by-step explanation:

Given,

Principal ( P ) = $ 6000

Amount ( A ) = $ 14550

Time ( T ) = 10 years

Rate ( R ) = ?

<u>Finding </u><u>the </u><u>Interest</u>

The sum of principal and interest is called an amount.

From the definition,

\boxed{ \sf{Amount =  \: Principal + Interest}}

plug the values

⇒\sf{14550 = 6000 + Interest}

Swap the sides of the equation

⇒\sf{6000 + Interest = 14550}

Move 6000 to right hand side and change its sign

⇒\sf{Interest = 14550 - 6000}

Subtract 6000 from 14550

⇒\sf{Interest = \: 8550 \: }

Interest = $ 8550

<u>Finding </u><u>the </u><u>rate </u>

{ \boxed{ \sf{Rate =  \frac{Interest \times 100}{Principal \times Time}}}}

plug the values

⇒\sf{ Rate = \frac{8550  \times 100}{6000 \times 10} }

Calculate

⇒\sf{Rate =  \frac{855000}{60000} }

⇒\sf{Rate = 14.25 \: \% \: }

Hope I helped!

Best regards!!

8 0
3 years ago
Item 29<br> Divide.<br> 6.8÷4<br> 6.8÷4
Ket [755]

Answer:

1.7

Step-by-step explanation:

You can just use a calculator to make the work easier.

7 0
2 years ago
Brandon graphed one line in a system of equations. The system has only one solution, (0,3). Which equations could also be part o
AlladinOne [14]

Answer:

y = 3 and y = x + 3

Step-by-step explanation:

The equation y = 3 whose graph is a line parallel to x-axis can be a part of the system.

Consider the equation y = x + 3.

Substitute (0, 3).

3 = 0 + 3

So, (0, 3) lies on y = x + 3 and this can also be a part of the system.

6 0
3 years ago
Read 2 more answers
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