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Wittaler [7]
3 years ago
8

hillary pays $9.25 of her monthly life insurance premium, and her employer covers the rest. if her monthly premium is $36.50, wh

at is the annual value to hillary of this benefit?
Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer:  Annual value to Hillary of this benefit is $327.

Step-by-step explanation:

Since we have given that

Amount of insurance premium paid by Hillary = $9.25

Total monthly premium = $36.50

Rest of the premium paid by her employer.

So, we need to find the annual value to Hillary of this benefit.

Amount of insurance premium paid by his employee is given by

\$36.50-\$9.25\\\\=\$27.25

Annual value would be

\$27.25\times 12\\\\=\$327

Hence, Annual value to Hillary of this benefit is $327.

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If a^2+1/a^2=79 where (a>0), find the value of a^3 +1/a^3
Nastasia [14]

Note the binomial expansion,

(<em>a</em> + 1/<em>a</em>)³ = <em>a</em> ³ + 3<em>a</em> + 3/<em>a</em> + 1/<em>a</em> ³

so

<em>a</em> ³ + 1/<em>a</em> ³ = (<em>a</em> + 1/<em>a</em>)³ - 3 (<em>a</em> + 1/<em>a</em>)

Similarly,

(<em>a</em> + 1/<em>a</em>)² = <em>a</em> ² + 2 + 1/<em>a</em> ²

We're given <em>a</em> ² + 1/<em>a</em> ² = 79, so

(<em>a</em> + 1/<em>a</em>)² - 2 = 79

(<em>a</em> + 1/<em>a</em>)² = 81

<em>a</em> + 1/<em>a</em> = ±9

but <em>a</em> > 0, so we ignore the negative solution.

Then

<em>a</em> ³ + 1/<em>a</em> ³ = 9³ - 3×9 = 702

7 0
3 years ago
How to solve inter quartile range of the given data set? 3,9,12,15,16,18,21
s2008m [1.1K]

Answer:

Step-by-step explanation:

3,9,12,(15),16,18,21

   

mean (middle number) = 15

Q1 = 9.....the middle number of the numbers before the mean

Q3 = 18...the middle number of the numbers after the mean

interquartile range (IQR) = Q3 - Q1 = 18 - 9 = 9 <==

7 0
3 years ago
Please help me on this I am stuck and I need help ASAP
k0ka [10]

Answer:

Grace is right

Step-by-step explanation:

3(2-x)-2(6x-8)=

so:

3×2 is 6

3×(-x) is -3x

-2×6x is -12x

-2×(-8) is 16 because -×- equals +

3(2-x)-2(6x-8)

=6-3x-12x+16

= -15x+22

6 0
3 years ago
Read 2 more answers
3, 30, 300, 3000 in explicit formula
Black_prince [1.1K]

Answer:

common ratio(r) = 10

place of the term = n

a = 3

then

sequence of 1 term = ar^(n-1)

= 3×10^(1-1)

= 3× 10^0

= 3×1

= 3

sequence of 2 term = ar^(n-1)

= 3×10¹

= 30

etc...

Step-by-step explanation:

lol id,k if this is right or not sooo yeah

6 0
3 years ago
the first four terms in an arithmetic sequence are $x y$, $x - y$, $xy$, and $x/y$, in that order. what is the fifth term?
olga2289 [7]

When the first four terms in an arithmetic sequence are x+y, x-y, xy and x÷y, then the fifth term will be \frac{123}{40}

A sequence of numbers in which every term (except the first term) is obtained by adding a constant number to the previous term is called an arithmetic sequence

In given arithmetic sequence

The first term = x+y

The second term = x-y

Then, the common difference= (x-y)-(x+y)

x-y-x-y= -2y

The third term = x-y+(-2y)

x-y-2y= x-3y

In question third term is given as xy

So both are equal

x-3y=xy

-3y=xy-x

-3y=x(y-1)

x= \frac{-3y}{y-1}

Similarly fourth term will be

x-3y-2y=x-5y

x-5y=\frac{x}{y}

substitute the value of x  in the equation

\frac{-3y}{y-1}-5y=\frac{\frac{-3y}{y-1} }{y} \\\frac{-3y}{y-1} -5y=\frac{-3}{y-1}\\ -3y-5y(y-1)=-3\\-3y-5y^{2}+5y=-3\\ -5y^{2} +2y+3=0\\5y^{2} -2y-3=0

Split the middle term and factorize it

5y^{2}+3y-5y-3=0\\y(5y+3)-1(5y+3)=0\\

(y-1)(5y+3)=0

y=1,\frac{-3}{5}

y cannot be equal to 1, because the third term and fourth term will become x.

y=\frac{-3}{5}

substitute the value of y in the equation of x
x=\frac{-3y}{y-1} \\=\frac{-3\frac{-3}{5} }{\frac{-3}{5}-1 }\\ =\frac{-9}{8}

The fifth term= \frac{x}{y}+(-2y)=\frac{x}{y} -2y

\frac{\frac{-9}{8} }{\frac{-3}{5} }-2\frac{-3}{5}  =\frac{123}{40}

Hence, when the first four terms in an arithmetic sequence are x+y, x-y, xy and x÷y, then the fifth term will be \frac{123}{40}

Learn more about Arithmetic sequence here

brainly.com/question/15412619

#SPJ4

5 0
2 years ago
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