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3 years ago

The bank agrees to give you a loan of

Mathematics

1 answer:

dusya [7]3 years ago

6 0

<h3>Answer: Monthly Payment = $12,083.34</h3>

========================================================

Explanation:

i = P*r*t ... simple interest formula

i = (1,500,000)*(0.03)*(15)

i = 675,000

The amount of interest you need to pay is $675,000

The total amount to pay back is 1,500,000+675,000 = 2,175,000 dollars.

Divide this over 180 months (since 15 years = 15*12 = 180 months) to get

(2,175,000)/180 = 12,083.3333333333

This rounds up to the nearest cent to get 12,083.34

I rounded up to make sure to clear the hurdle. Note how

180*(12,083.33) = 2,174,999.4

which is a bit short of the 2,175,000 mark we're after

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B. 2(3x - 5) = 2x + 6

Leto [7]

Answer:

x = 4

Step-by-step explanation:

Given

2(3x - 5) = 2x + 6 ( divide both sides by 2 )

3x - 5 = x + 3 ( subtract x from both sides )

2x - 5 = 3 ( add 5 to both sides )

2x = 8 ( divide both sides by 2 )

x = 4

8 0

3 years ago

Read 2 more answers

a box is 24cm by 15cm by 10cm. this box will packed with smaller boxes of paper clips which are 6cm by 3cm by 5cm . what is the

vichka [17]

Answer: 40

Align in this pattern:

Bigger box : 24 : 15 : 10

smaller box: 6 : 3 : 5

counts: 4: 5 : 2

( multiply 4*5*2 = 40)

7 0

3 years ago

What is the angle of rotation of the minute hand of a clock moving from 6:10 to 7:00?

igomit [66]

Because the movement of the minute han of a clock is clockwise ↓, then the angle of rotation will be negative.

1) we calculate the number of minutes from 6:10 to 7:00

number of minutes=7:00-6:10=6:60 - 6:10=50 minutes.

2)
We know that:
60 minutes of a clock have turned -360º
Then:

60 minutes of a clock------------------------- -360º
50 minutes of a clock------------------------- x

x=(50 minutes of a clock * (-360º))/60 minutes of a clock=-300º

Answer: A. -300º

5 0

4 years ago

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In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$