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1 year ago

John takes out a loan for $ 10 , 000 at a simple interest rate of 5 % to be paid back in 36 monthly installments.

Mathematics

1 answer:

andrew-mc [135]1 year ago

6 0

The monthly principal and interest payment John will pay monthly is $3,750.04

<h3>What is the amount of principal and interest repayment?</h3><h3 />

Given:

P = 10,000
i = 5% / (12 months) = 0.375
L = 36 months

$M = 10000 \times \frac{0.375}{ {1 - (1 + 0.375)}^{ - 36} }$

$M = 10000 \times \frac{0.375}{ {1 - (1.375)}^{ - 36} }$

$M = 10000 \times \frac{0.375}{ {1 - (0.00001049791)} }$

$M = 10000 \times \frac{0.375}{ 0.99998950208 }$

$M = 10000 \times 0.37500393675$

$M = 3750.03936758$

Approximately,

M = $3,750.04

Therefore, John will be be repaying $3,750.04 monthly including principal and interest

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The hour hand of a clock is 3 feet long. How many feet does

dem82 [27]

Answer:

2π feet

Step-by-step explanation:

9:30 PM to 1:30 AM is 4 hours.

It is 1/3 of a complete revolution.

A full 12 hours would be 2×r×π=2×3×π

1/3 of that is:

1/3×2×3×π=2π

≈6.283185307 feet

3 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

Mike bought a soft drink for 2 dollars and 7 candy bars. He spent a total of 23 dollars. How much did each candy bar cost? Write

hram777 [196]

21 divided by 7 is 3 so each candy bar equals 3 dollars

4 0

3 years ago

Read 2 more answers

Find what x is 17+7x=x+5

marusya05 [52]

Step-by-step explanation: