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Gelneren [198K]
3 years ago
5

Please help me with 5

Mathematics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

Fourth chice, 3/5

Step-by-step explanation:

Add all the numbers together, 12 + 3 + 5 = 20

There's 12 green cubes out of the 20 cubes so it's 12/20 which is 3/5

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Sergio's internet provider charges its customers $9 per month plus 4¢ per minute of on-line usage. Sergio received a bill from t
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$81.40-$9.00=$72.40
$72.40/.4 (cents)= 181 minutes 
8 0
3 years ago
Robert invests $1100 at an annual interest rate of 8% for the period of 1 year. What equation shows the relationship between pri
Sholpan [36]

Answer:

88 dollars

The equation would be:

prt=i

(1)(.08)(1100)=88

Hope this helps! ☺

Step-by-step explanation:

8 0
3 years ago
Which is an equation of a line parallel to -2/3x + 1/5y = 4
IrinaVladis [17]

Answer:

y = 3.3x + 20

Step-by-step explanation:

-2/3x +1/5y = 4   y = mx + b

-2/3x + 1/5y = 4

-2/3                -2/3

1/5y = 2/3 + 4

1/5  / 2/3 = 3.3

1/5 / 4 = 20

y = 3.3x + 20

8 0
3 years ago
6. If the net investment function is given by
Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated  capital for the period.

  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

  • (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0
2 years ago
Cuanto es <br> (99+24)-47*(920+41)
Advocard [28]

Answer:

-45044

Step-by-step explanation:

8 0
3 years ago
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