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zysi [14]
3 years ago
13

Consider a student loan of ​$20,000 at a fixed APR of 12​% for 15 years. a. Calculate the monthly payment.​(Do not round until t

he final answer. Then round to the nearest cent as​ needed.) b. Determine the total amount paid over the term of the loan. ​(Round to the nearest cent as​ needed.) c. Of the total amount​ paid, what percentage is paid toward the principal and what percentage is paid for interest. ​(Round to the nearest tenth as​ needed.)

Mathematics
1 answer:
joja [24]3 years ago
6 0

Answer:

a. $240.03

b. $43,205.04

c. Principal = 46.29%

   Interest = 53.71%

Step-by-step explanation:

The computation is shown below:

a. For computing the monthly payment we need to apply the PMT formula i.e to be shown in the attachment below

Given that,  

Present value = $25,000

Future value or Face value = 0

RATE = 12% ÷ 12 months = 1%

NPER = 15 years × 12 months = 180 months

The formula is shown below:  

= PMT(RATE;NPER;-PV;FV;type)  

The present value come in negative  

So, after applying the above formula the monthly payment is $240.03

b. Now the total amount paid is

= Monthly payment × time period

= $240.03 × 180 months

= $43,205.04

c. Now the percentage is

For principal

= $25,000 ÷ $43,205.04 × 100

= 46.29%

And, for the interest, it is

= 100% - 46.29%

= 53.71%

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A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performanc
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Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

<em><u /></em>

<em><u>Using Two-way ANOVA method</u></em>

Given problem

<em><u>Observation              I          II       III          Row total (xr)</u></em>

A                              18 21 20            59

B                            24 26 27             77

C                            30 29 34             93

D                            22 25 24            71

<u>E                            20 23 24           63                      </u>

Col total (xc)             114 124 129        367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

=9196.3333→(C)

(∑x)²/n

=(367)²/15

=134689/15

=8979.2667→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

=217.0667

Sum of squares between columns

SSC=∑x²c/r-(∑x)²/n

=(B)-(D)

=9002.6-8979.2667

=23.3333

Sum of squares Error (residual)

SSE=SST-SSR-SSC

=253.7333-217.0667-23.3333

=13.3333

<u>ANOVA table</u>

Source                 Sums         Degrees      Mean Squares

of Variation       of Squares   of freedom

<u>                               SS                 DF              MS       F p-value</u>

B/ w     SSR=217.0667              4 MSR=54.2667    32.56 0.0001

rows

B/w     SSC=23.3333         c-1=2 MSC=11.6667        7 0.01

columns

<u>Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667                  </u>

<u>Total SST=253.7333 rc-1=14                                                        </u>

Conclusion:

<u> 1. F for between Rows</u>

The critical region for F(4,8) at 0.05 level of significance=3.8379

The calculated F for Rows=32.56>3.8379

Therefore H0 is rejected

<u>2. F for between Columns</u>

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

<u><em>Part B:</em></u>

To analyze the data for completely  randomized designs click on anova two factor without replication  in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF             Sum                  Mean           F Statistic

<u>                 (df1,df2)    of Square (SS) Square (MS)                    P-value</u>

Factor A       1 1496.5444 1496.5444 769.6514          0.001297

Rows

Factor B -     2 19.4444           9.7222               5                  0.1667

Columns

Interaction

AB               2    3.8889   1.9444        0.1013         0.9045

<u> Error     12   230.4            19.2                                           </u>

<u>Total 17 1750.2778 102.9575                                                         </u>

<u />

<u>Factor - A- Rows</u>

Since p-value < α, H0 is rejected.

<u>Factor - B- Columns</u>

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

<u>Interaction AB</u>

Since p-value > α, H0 can not be rejected.

<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

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