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blsea [12.9K]
2 years ago
6

A payday loan company charges a $35 fee for a $350 payday loan that will be repaid in 19 days. Treating the fee as interest paid

, what is the equivalent annual interest rate?
Mathematics
1 answer:
Naily [24]2 years ago
6 0

Answer:

  192.1%

Step-by-step explanation:

The formula for the amount of interest is ...

  i = Prt . . . . . where i is the interest amount, P is the principal, r is the annual rate, and t is the number of years. (Here, we have a fraction of a year.)

Solving for the rate, we have ...

  r = i/(Pt)

For "exact" interest, we use 365 days in the year. So, for a time of 19 days of 365, the interest rate is ...

  r = $35/($350·19/365) ≈ 1.921 = 192.1%

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Effectus [21]
The correct answer is:  [C]:  " \frac{7}{18}" .
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Explanation:
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Given:
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  2 / (9x) = 4 / 7 ;  solve for "x" ;
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Cross-multiply:

        →   (9x)*4  = 2 * 7 ; 

        →   36x = 14 ; 

Divide each side of the equation by "36" ; 
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        →   36x / 36 = 14 / 36 ; 

                   x = 14/ 36 ; 

                  → x = 14/36 = (14÷2) / (36÷2) = 7/18 ; 
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The answer is:  " \frac{7}{18}" ;

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A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco
sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

            n = sample of cars = 600

So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

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<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

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