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saveliy_v [14]
3 years ago
12

Amanda just took out a loan for $950 at a 7.2% APR, compounded monthly, to

Mathematics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

C. 16 months faster

Step-by-step explanation:

You can solve this using a Financial calculator( TI BA II plus in this case)

First, find number of months if recurring monthly payment is $38.50;

Amount borrowed ; PV = -950

Monthly rate ; I/Y  = 7.2%/12 = 0.6%

Monthly payment; PMT = 38.50

Future value ; FV = 0

Total duration;  press CPT, N = 26.785  OR 27 months

Next, find number of months if recurring monthly payment is $93;

Amount borrowed ; PV = -950

Monthly rate ; I/Y  = 7.2%/12 = 0.6%

Monthly payment; PMT = 93

Future value ; FV = 0

Total duration;  press CPT, N = 10.573 OR 11 months

Difference = 27 -11 = 16

Therefore, she would be able to pay off the loan 16 months faster.

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4y = 42 - 3y<br><br> Can you help me find y?
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4y = 42 - 3y

First, add '3y' to both of the sides.
4y + 3y = 42
Second, add '4y + 3y' to get '7y'.
7y = 42
Third, divide both sides by '7'.
y =  \frac{42}{7}
Fourth, how many times does 7 go into 42? 42 ÷ 7 = '6'.
y = 6

Answer: y = 6

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3 years ago
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What is the value of x in the equation 3x-4y=65, when y=4?<br> WN
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Answer:

x=27

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3x -4y =65

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3x -4*4 =65

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Divide each side by 3

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Evaluate 3|-4|<br> 2<br> -2<br> 1
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You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a r
SIZIF [17.4K]

Answer:

The critical value that corresponds to a confidence level of 99% is, 2.58.

Step-by-step explanation:

Consider a random variable <em>X</em> that follows a Binomial distribution with parameters, sample size <em>n </em>and probability of success <em>p</em>.

It is provided that the distribution of proportion of random variable <em>X, </em>\hat p, can be approximated by the Normal distribution.

The mean of the distribution of proportion is, \mu_{\hat p}=\hat p

The standard deviation of the distribution of proportion is, \sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}.

Then the confidence interval for the population proportion <em>p</em> is:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }

The confidence level is 99%.

The significance level is:

\alpha =1-\frac{Confidence\ level}{100}=1-\frac{99}{100}=1-0.99=0.01

Compute the critical value as follows:

z_{\alpha /2}=z_{0.01/2}=z_{0.005}

That is:

P(Z>z)=0.005\\P(Z

Use the <em>z</em>-table for the <em>z-</em>value.

For <em>z</em> = 2.58 the P (Z < z) = 0.995.

And for <em>z</em> = -2.58 the P (Z > z) = 0.005.

Thus, the critical value is, 2.58.

7 0
4 years ago
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