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crimeas [40]
3 years ago
4

In many cases, lenders allow homeowners to include their homeowners insurance premium with their monthly mortgage payment. Tim’s

home is worth $279,500. If his homeowners insurance premium is $0. 33 per $100, how much is added to his monthly mortgage payment for insurance? a. $7. 69 b. $76. 86 c. $92. 35 d. $922. 35 Please select the best answer from the choices provided A B C D.
Mathematics
1 answer:
salantis [7]3 years ago
3 0

The monthly mortgage payment for insurance is $76.86. Option B shows the correct payment for insurance.

<h3>How do you calculate the monthly mortgage payment for insurance?</h3>

Given that Tim’s home is worth $279,500 and his homeowners insurance premium is $0.33 per $100.

The premium is on yearly basis. Then, the premium for a month is given as,

1 month premium on $100 = \dfrac {\$0.33 }{12}

1 month premium on $100 = $0.0275

1 month premium on $1 = \dfrac {\$0.0275}{100}

Now the premium is,

1 month premium on $279500 = \dfrac {\$0.0275}{100} \times \$279500

1 month premium on $279500 = $76.86

Hence we can conclude that the monthly mortgage payment for insurance is $76.86. Option B shows the correct payment for insurance.

To know more about premium, follow the link given below.

brainly.com/question/4304080.

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Answer:

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Factor completly 4x^2 +8x -5
DaniilM [7]

∑ Hey, jillianwagler ⊃

Answer:

\left(2x-1\right)\left(2x+5\right)

Step-by-step explanation:

<u><em>║Given info║:</em></u>

<em>Factor completely: </em>4x^2 +8x -5

<u><em>Solution~:</em></u>

<em>Breaking the expression~:  </em>\left(4x^2-2x\right)+\left(10x-5\right)

<em>Factor out 2x from 4x² -2x : 2x(2x-1)</em>

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<em>Put together: </em>=2x\left(2x-1\right)+5\left(2x-1\right)

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2 years ago
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The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

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and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

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