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3 years ago

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 298.87.

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

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 ≈ 46.15 years.

Learn more investment function here:

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6 0

2 years ago

Cuanto es (99+24)-47*(920+41)

Advocard [28]

Answer:

-45044

Step-by-step explanation:

8 0

3 years ago