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Katyanochek1 [597]
3 years ago
11

If Samantha can pay off her loan in 36 months at a 10% interest rate rather than in 48 months at a 12% interest rate, how much m

oney will she save in interest charges on her $6,000 loan?
Mathematics
1 answer:
Ymorist [56]3 years ago
7 0
Hi there
The simple interest formula is
I=prt
I interest changes
P amount of the loan 6000
R interest rate
T time( number of months/12 months)

The interest in 36 months at a 10%
I=6,000×0.1×(36÷12)=1,800
The interest in 48 months at a 12%
6,000×0.12×(48÷12)=2,880
she will save
2,880−1,800=1,080

Good luck!
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nikdorinn [45]

Answer:

See explanation

Step-by-step explanation:

(x^2+9x+10)(x-2)= \\\\x^3-2x^2+9x^2-18x+10x-20= \\\\x^3+7x^2-8x-20

Jimmy is incorrect.

Using long division, you can find that the real answer.

First, subtract x^2(x-2) from the original expression, leaving you with 9x^2+2x-40.

Next, subtract 9x(x-2) from the expression, leaving you with 20x-40. Finally, subtract 20(x-2) from the expression, leaving you with a remainder of 0. This means that the real quotient is x^2+9x+20. Hope this helps!

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3 years ago
Evaluate: 10x when x=10<br> The answer is_______( just type in your final number answer)​
Kisachek [45]

Answer:

100

Step-by-step explanation:

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True or false<br> The solution set of 2x+5=x-3is {-8}
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Answer:

True

Step-by-step explanation:

2x+5=x-3 is (-8)

This is the correct answer

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4 years ago
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The number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.
Alchen [17]

Answer:

Probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is 0.89435.

Step-by-step explanation:

We are given that the number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.

Also, sample of 49 adults is given.

<em>Let X = number of hours spent per week on household chores</em>

So, assuming data follows normal distribution; X ~ N(\mu=28,\sigma^{2}=7^{2})

The z score probability distribution for sample mean is given by;

               Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = mean hours = 28

            \sigma = standard deviation = 7 hours

            n = sample of adults = 49

<em>Let </em>\bar X<em> = sample mean hours spent per week on household chores</em>

So, probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is given by =P(\bar X > 26.75 hours)

    P(\bar X > 26.75 hours) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{26.75-28}{\frac{7}{\sqrt{49} } } ) = P(Z > -1.25) = P(Z < 1.25)

                                                                     = 0.89435  {using z table}

Therefore, probability that the mean hours spent per week on household chores is more than 26.75 is 0.89435.

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