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2 years ago

6

Bill was looking at 13 different colors to paint his bedroom he liked about 2/3 of the colors. about how many colors did he like

1 answer:

Jet001 [13]2 years ago

8 0

He liked about 8-9 colors because all you have to do is subtract 33% which equals 8.71

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Twenty one children and three teachers went to the movies. The total bill was $105. The price for

MrRissso [65]

Answer:

c = price of child ticket = $4.25

a = price of adult ticket = $4.25 + $1.00 = $5.25

Step-by-step explanation:

A child ticket costs c and an adult ticket costs a = c + 1.

21 children pay c dollars each for admission, and

3 teachers pay c + 1 dollars each for admission.

Thus,

21c + 3(c + 1) = $105, and

21c + 3c + 3 = $105, so that:

24c = $102

c = price of child ticket = $4.25

a = price of adult ticket = $4.25 + $1.00 = $5.25

3 0

2 years ago

A small company has 20,000 shares. Anastasia owns 400 of these shares. The company decides to split its shares. What is Anastasi

Pani-rosa [81]

The answer is the B one or 2%.

8 0

3 years ago

Please help me will give brainlest to correct answer and simple/small explanation

BabaBlast [244]

Answer:

the measure of the angle is 72

x=36

Step-by-step explanation:

we can make an algebraic equation using the info given

2x+80=152

2x=72

x=36

the measure of the angle is 72

3 0

2 years ago

Read 2 more answers

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

2 years ago

Evaluate the expression and enter your answer 9+|9+2|

Jet001 [13]

9+|9+2|

Add 9 and 2

9+11

Add 9 and 11

20

3 0

2 years ago