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ANTONII [103]
3 years ago
12

How do you get $3,960 as answer for the question, Misty deposits $3,600 in a bank account that pays 2% simple interest. If she d

oesn't change the principal, what will her balance be in 5 years?
Mathematics
1 answer:
Whitepunk [10]3 years ago
3 0
Because 2 percent as a decimal is .02 so times that by the annual amount 5 years and you'd get 10. Then you times it with your original amount 3600 and get 360 . then you add them together 3600 and 360 and get 3960
You might be interested in
Write the polynomial 8x − 27 + 2x2y − y2 + 7y in standard form.
Westkost [7]
Your answer is 2x^2y-y^2+8x+7y-27
3 0
4 years ago
Counting bit strings. How many 10-bit strings are there subject to each of the following restrictions? (a) No restrictions. The
-BARSIC- [3]

Answer:

a) With no restrictions, there are 1024 possibilies

b) There are 128 possibilities for which the tring starts with 001

c) There are 256+128 = 384 strings starting with 001 or 10.

d) There are 128  possiblities of strings where the first two bits are the same as the last two bits

e)There are 210 possibilities in which the string has exactly six 0's.

f) 84 possibilities in which the string has exactly six O's and the first bit is 1

g) 50 strings in which there is exactly one 1 in the first half and exactly three 1's in the second half

Step-by-step explanation:

Our string is like this:

B1-B2-B3-B4-B5-B6-B7-B8-B9-B10

B1 is the bit in position 1, B2 position 2,...

A bit can have two values: 0 or 1

So

No restrictions:

It can be:

2-2-2-2-2-2-2-2-2-2

There are 2^{10} = 1024 possibilities

The string starts with 001

There is only one possibility for each of the first three bits(0,0 and 1) So:

1-1-1-2-2-2-2-2-2-2

There are 2^{7} = 128 possibilities

The string starts with 001 or 10

There are 128 possibilities for which the tring starts with 001, as we found above.

With 10, there is only one possibility for each of the first two bits, so:

1-1-2-2-2-2-2-2-2-2

There are 2^{8} = 256 possibilities

There are 256+128 = 384 strings starting with 001 or 10.

The first two bits are the same as the last two bits

The is only one possibility for the first two and for the last two bits.

1-1-2-2-2-2-2-2-1-1

The first two and last two bits can be 0-0-...-0-0, 0-1-...-0-1, 1-0-...-1-0 or 1-1-...-1-1, so there are 4*2^{6} = 256 possiblities of strings where the first two bits are the same as the last two bits.

The string has exactly six o's:

There is only one bit possible for each position of the string. However, these bits can be permutated, which means we have a permutation of 10 bits repeatad 6(zeros) and 4(ones) times, so there are

P^{10}_{6,4} = \frac{10!}{6!4!} = 210

210 possibilities in which the string has exactly six 0's.

The string has exactly six O's and the first bit is 1:

The first bit is one. For each of the remaining nine bits, there is one possiblity for each.  However, these bits can be permutated, which means we have a permutation of 9 bits repeatad 6(zeros) and 3(ones) times, so there are

P^{9}_{6,3} = \frac{9!}{6!3!} = 84

84 possibilities in which the string has exactly six O's and the first bit is 1

There is exactly one 1 in the first half and exactly three 1's in the second half

We compute the number of strings possible in each half, and multiply them:

For the first half, each of the five bits has only one possibile value, but they can be permutated. We have a permutation of 5 bits, with repetitions of 4(zeros) and 1(ones) bits.

So, for the first half there are:

P^{5}_{4,1} = \frac{5!}{4!1!} = 5

5 possibilies where there is exactly one 1 in the first half.

For the second half, each of the five bits has only one possibile value, but they can be permutated.  We have a permutation of 5 bits, with repetitions of 3(ones) and 2(zeros) bits.

P^{5}_{3,2} = \frac{5!}{3!2!} = 10

10 possibilies where there is exactly three 1's in the second half.

It means that for each first half of the string possibility, there are 10 possible second half possibilities. So there are 5+10 = 50 strings in which there is exactly one 1 in the first half and exactly three 1's in the second half.

5 0
3 years ago
What would it be if -3/4×-3/1=
Finger [1]
2 1/4 (2.25) is your answer
5 0
4 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
BabaBlast [244]

Answer: 8

Step-by-step explanation:

Pythagorean theorem - a^2+b^2=c^2

a^2+15^2=17^2

a^2+225=289

(subtract 225 from both sides)

a^2=64

(sq root both sides)

a=8

6 0
3 years ago
What is the slope of a line parallel to the line with this equation:
Nutka1998 [239]

The slope of two parallel lines are equal.

Let us find slope of given equation.

The general slope intercept form is given by :

y=mx+c

where m is slope

our equation is :

5x-15y=30

Now we have to convert our equation into slope intercept form.

let us try to isolate y on the left side.

taking 5x to the right side we have ,

-15y=-5x+30

dividing both sides by -15,

y=\frac{x}{3} -2

coefficient of x is slope,

so slope is 1/3

slope of line parallel to it is also same that is 1/3 .

Correct option is D.

7 0
3 years ago
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