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Brut [27]
3 years ago
5

A vehicle covers 360 miles in 5.5 hours. What is the average speed of the vehicle (in miles per hour) to the nearest tenth.

Mathematics
2 answers:
AnnZ [28]3 years ago
3 0
I think it is 65.45 mph cause 360 divided by 5.5 = 65.45

Nikitich [7]3 years ago
3 0
<h2>Answer:</h2>

The Average speed of the vehicle to the nearest tenth is:

                  65.4 miles per hour.

<h2>Step-by-step explanation:</h2>

It is given that the total distance covered by the vehicle is: 360 miles.

Also, total time taken by vehicle is: 5.5 hours.

Now we are asked to find the average speed of the vehicle.

We Know that average speed is the ratio of total distance traveled to total time taken.

i.e.

Average\ Speed=\dfrac{Total\ distance}{Total\ time}

i.e.

Average\ Speed=\dfrac{360}{5.5}\\\\Average\ Speed=65.4545\ \text{miles\ per\ hour}

Hence, Average speed is: 65.4 miles per hour.

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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
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