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Dmitry_Shevchenko [17]
3 years ago
9

A printing business employs 35 people. The employer offers the employees an insurance package that costs the company $2,170. The

employer has been investigating various plans. A new plan would cost $59 per employee plus a one-time $70 sign up fee (not per employee). Is the new plan a better deal?
Mathematics
2 answers:
7nadin3 [17]3 years ago
5 0

Answer:

New plan better

Step-by-step explanation:

We need to compare the cost of the first plan, $2170, to the cost of this  new plan. Fot this, we need to find the cost of the new plan. We need an equation for this.

For the new plan there is a fixed cost of $70, i.e, no matter how many employees there will be, the firm would have to pay $70. Then, if it has 1 employee, would face a cost of $70 + $59, where this $59 is the cost per employee. If the firm has 2 employees the cost will be $70 + $59 + $59 = $70 + 2*($59). If it hires 3 employees: $70 + $59 +$59 +$59 = $70 + 3*($59). And so forth for 4, 5, 6 employees.

Then, for the 35 employees the firm has the cost is:

$70 + 35*($59) = $2135

As $2135 < $2170, this new plan is better

Alex3 years ago
5 0

Answer:yes

Step-by-step explanation:

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Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

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What is the probability of rolling a dice and getting either a 1 or 2 and flipping a coin and getting a heads?
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Step-by-step explanation:

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I need help with #61 to #64 ASAP please and thank you …. Could someone please help me with it… I need to get it done ASAP
aivan3 [116]

Problem 61

The nth triangular number is

T(n) = n(n+1)/2

I'll rewrite this into

T(n) = 0.5n(n+1)

The triangular number right after this is

T(n+1) = 0.5(n+1)(n+2)

I replaced every n with n+1 and simplified

Let's see what we get when we add up the two expressions

T(n) + T(n+1)

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0.5n^2+0.5n + 0.5(n^2+3n+2)

0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1

n^2+2n+1

(n+1)^2

This shows that the sum of any two consecutive triangular numbers results in a square number

Here's a few examples

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  • 1+3 = 4
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  • 6+10 = 16
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Note each sum is a perfect square, which visually would plot out a square figure.

For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}

<h3>Answer: Square number</h3>

==========================================================

Problem 64

Let's say we go with n = 5.

This means,

T(n) = 0.5n(n+1)

T(n-1) = 0.5(n-1)(n-1+1)

T(n-1) = 0.5n(n-1)

T(5-1) = 0.5*5(5-1)

T(4) = 10

This says that when n = 5, the 4th triangular number is 10

Triple that result and add on n = 5

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Through a bit of trial and error, you should find it's a pentagonal number

Pentagonal numbers are of the form n(3n-1)/2

If you plugged n = 5 into that, it leads to 35

n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35

The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)

<h3>Answer: Pentagonal number</h3>

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