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2 years ago

7

Pam has monthly mortgage payment of $750. Her annual property taxes are $1200,her PMI is $55/month, and her homeowners insurance

is $800 per year. How much is deposited into Pam’s escrow account each month?

1 answer:

Alenkinab [10]2 years ago

7 0

Answer:

$100 + $66.67 + $55 = $221.67/month, which is the amount that must be deposited into Pam's escrow account each month....

Step-by-step explanation:

Pam's annual taxes are $1200 per year. It means her escrow each month needs to have 1/12 of that amount. 1200/12 = $100 per month

Her homeowner's insurance is $800 per year, that's 1/12 of that needs to be in her escrow account each month:

$800 / 12, or $66.67/month.

And PMI is $55/month

Add these three together:

$100 + $66.67 + $55 = $221.67/month, which is the amount that must be deposited into Pam's escrow account each month....

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Error Analysis: Kayla said the remainder of x^3+ 2x^2 -4x + 6 divided by

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9514 1404 393

Answer:

C) No, Kayla is not correct, because the remainder is -49 not 149

Step-by-step explanation:

The remainder from division by x+5 can be found by evaluating the polynomial at x=-5, the value of x that makes the divisor zero.

Evaluation is often easier when the function is written in Horner form.

((x +2)x -4)x +6

For x=-5, this is ...

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2 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

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