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notsponge [240]
3 years ago
15

An art teacher has a total of 7/8 pound of clay.The teacher puts 1/16 pound of clay at each workstation. The teacher sets up an

equal number of workstations in each of 2 classrooms. How many workstations does the teacher set up in each of the classrooms? Show your work
Mathematics
1 answer:
Pie3 years ago
7 0

Answer:

The teacher establishes 7 work stations in each of the classrooms.

Step-by-step explanation:

Each work station has 1/16 pounds of clay.

There are two classrooms with the same number of workstations.

The total amount of clay is 7/8

Then, we look for a number of work stations per classroom x, which multiplied by the number of classrooms (2) and multiplied by the amount of clay per work station (1/16) is equal 7/8.

Then we propose the following equation

\frac{1}{16}(2x) =\frac{7}{8}\\\\ \frac{2}{16}x = \frac{7}{8}\\\\ x = \frac{16 * 7}{8 * 2}\\\\ x = 7

The teacher establishes 7 work stations in each of the classrooms.

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You find an interest rate of 10% compounded quarterly. Calculate how much more money you would have in your pocket if you had us
Elena-2011 [213]

Answer:

see the explanation

Step-by-step explanation:

we know that    

step 1

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

in this problem we have  

r=10\%=10/100=0.10\\n=4  

substitute in the formula above

A=P(1+\frac{0.10}{4})^{4t}  

A=P(1.025)^{4t}  

Applying property of exponents

A=P[(1.025)^{4}]^{t}  

A=P(1.1038)^{t}  

step 2

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

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  

e is the mathematical constant number

we have  

r=10\%=10/100=0.10  

substitute in the formula above

A=P(e)^{0.10t}  

Applying property of exponents

A=P[(e)^{0.10}]^{t}  

A=P(1.1052)^{t}  

step 3

Compare the final amount

P(1.1052)^{t} > P(1.1038)^{t}

therefore

Find the difference

P(1.1052)^{t} - P(1.1038)^{t} ----> Additional amount of money you would have in your pocket if you had used a continuously compounded account with the same interest rate and the same principal.

3 0
3 years ago
What operation should be used to solve the equation 153=g+45
Advocard [28]

Answer:

Subtraction

Step-by-step explanation:

Subtraction

153 - 45 = g + 45 - 45

108 = g

3 0
3 years ago
What is y=-3/4+2 in standard form
Elenna [48]
Right answer is
Y=5/4
5 0
3 years ago
A customer borrowed $2000 and then a further $1000 both repayble in 12 months. What would he have saved if he had taken out one
RSB [31]

A customer borrowed $2000 and then a further $1000 both repayble in 12 months. What would he have saved if he had taken out one loan for $3000 repayable in 12 months?

He took two different loans, it charged him loan processing fee twice, two-time documentation process, and of course, extra time spent for second loan. Instead, he could take single loan of $3000 with one-time processing fee, one-time documentation process, and time-saving also.

8 0
3 years ago
Dan bought a truck for $29,800. The value of the truck depreciated at a constant rate per year. The table shows the value of the
leonid [27]

Answer:

A) f(t) = 29,800(0.89)^t

Step-by-step explanation:The truck value decreases by 11% each year.

The present value of the truck was $29,800.

29,800 x 0.11 = 3,728 dollars lost from value.

29,800 - 3,728 = 26,522, the value of the truck after one year.

26,522 x 0.11 = 2,917.42 cash lost from value.

26,522 - 3,728 = 23,604.58, the value of the truck after two years.

Therefore, the answer is A) f(t) = 29,800(0.89)^t.

7 0
2 years ago
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