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Svetllana [295]
2 years ago
5

Hi can someone please help me with this i will give you five stars!!

Mathematics
1 answer:
tino4ka555 [31]2 years ago
3 0

Step-by-step explanation:

Let's look at what we know. We know that...

P=2000

r=0.04 (Change 4% to a decimal)

t=7 (25 years minus 18 years equals 7)

n=1

Since we are compounding each year, we need to use this equation: P(1+\frac{r}{n} )^{nt}

Now just plug the numbers in: (Answer to #1:) 2000(1+\frac{0.04}{1} )^{(1*7)}

This equals 2631.86

When we round, this equals to $2,632. (Answer to #2).

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Where do you see rotation in our lives/world?
MArishka [77]

Answer:

There's a rotation in our schedule if things we do everyday.

Step-by-step explanation:

Hope this helps :D

4 0
3 years ago
Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
3(x+3)+3x=15<br><br> Solution and check<br><br> Shown in attachment below
OLga [1]
3x + 9 + 3x =15
6x + 9 = 15
- 9 = -9
6x = 6
/6 /6

X =1
6 0
3 years ago
17. Let f(x) = 2x2 + 5x + 6 and g(x) = 3x2 + 4x – 10.<br> a Find f(x) + g(x)<br> b. Find f(x) – g(x)
nalin [4]

Step-by-step explanation:

f(x)+g(x)=(2x^2 + 5x +6) + (3x^2 +4x - 10)

Combine like terms

5x^2 + 9x -4

f(x)-g(x)=(2x^2 + 5x +6) - (3x^2 +4x - 10)

Distribute the negative

(2x^2 + 5x +6) + (-3x^2 -4x + 10)

Combine like terms

-x^2 +x +16

7 0
3 years ago
Read 2 more answers
Will is twice as old as Jill. Three years ago, Jill's age was two fifths of Will's age then. How old is each now?
dimulka [17.4K]
"Will is twice as old as Jill."

                       Jill's age . . . . .   J
                       Will's age . . . . 2J .

"Three years ago . . .

                       Jill's age then . . . . .   J - 3
                       Will's age then . . . . 2J - 3

". . . Jill's age then was 2/5 of Will's age then."

                                         J - 3  =  (2/5) (2J - 3)
Multiply
each side by  5 :          5J - 15  =  2 (2J - 3)

Divide
each side by  2 :          2.5 J - 7.5  =  2J - 3 

Subtract  2J
from each side:            0.5 J - 7.5  =      -3

Add  7.5
to each side:                0.5 J           =     4.5

Multiply
each side by  2 :                J            =       9

Jill is  9  y.o. now.
Will is 18 y.o. now.

4 0
3 years ago
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