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Karolina [17]
4 years ago
11

Harold and his friend both enjoy running long distances.Yesterday Harold ran 8 2/4 miles and his friend ran 4 3/4 miles. How man

y more miles did Harold run than his friend. Answer as a mixed number
Mathematics
1 answer:
andriy [413]4 years ago
4 0

Answer:

Harold ran 3\frac{3}{4}\ miles more than his friend.

Step-by-step explanation:

Given:

Number of miles Harold ran = 8\frac{2}{4}\ miles

8\frac{2}{4}\ miles can be Rewritten as \frac{34}{4}\ miles

Number of miles Harold ran = \frac{34}{4}\ miles

Number of miles his friend ran = 4\frac{3}{4}\ miles

4\frac{3}{4}\ miles can be Rewritten as \frac{19}{4}\ miles

Number of miles his friend ran = \frac{19}{4}\ miles

We need to find number of miles Harold ran more than his friend.

Solution:

Now we can say that;

To find the number of miles Harold ran more than his friend we need to subtract Number of miles his friend ran from Number of miles Harold ran.

framing in equation form we get;

the number of miles Harold ran more than his friend = \frac{34}{4}-\frac{19}{4} =\frac{34-19}{4} = \frac{15}{4}\ miles

\frac{15}{4}\ miles can be Rewritten as 3\frac{3}{4}\ miles

the number of miles Harold ran more than his friend = 3\frac{3}{4}\ miles

Hence Harold ran 3\frac{3}{4}\ miles more than his friend.

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emmainna [20.7K]

Answer: 432 units²

Step-by-step explanation:

The figure is composed by two trapezoids.

The formula for calculate the area of a trapezoid is:

A=\frac{h}{2}(B+b)

Where "B" is the larger base, "b" is the smaller base and "h" is the height.

Let be A_f the area of the figure, A_1 the area of the trapezoid on the left and A_2 the area of the trapezoid of the right. Then the area of the figure will be:

 A_f=A_1+A_2

A_f=\frac{h_1}{2}(B_1+b_1)+\frac{h_2}{2}(B_2+b_2)

Substituting values, you get:

A_f=\frac{16units}{2}(25units+4units)+\frac{10units}{2}(25units+15units)=432units^2

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Step-by-step explanation:

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You saved $20,000.00 and want to diversify your monies. You invest 45% in a Treasury bond for 3 years at 4.35% APR compounded an
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Compound Interest

A total of $20,000 is invested in different assets.

45% is invested in a Treasury bond for 3 years at 4.35 APR compounded annually.

For this investment, the principal is P = 0.45*$20,000 = $9,000.

The compounding period is yearly, thus the interest rate is:

i = 4.35 / 100 = 0.0435

The duration (in periods) is n = 3

Calculate the final value with the formula:

M=P_{}(1+i)^n

Substituting:

\begin{gathered} M=\$9,000_{}(1+0.0435)^3 \\ M=\$9,000\cdot1.136259062875 \\ M=\$10,226.33 \end{gathered}

The second investment is a CD at 3.75% APR for 3 years compounded annually. The parameters for the calculations are as follows:

P = 15% of $20,000 = $3,000

i = 3.75 / 100 = 0.0375

n = 3

Calculating:

\begin{gathered} M=\$3,000_{}(1+0.0375)^3 \\ M=\$3,000\cdot1.116771484375 \\ M=\$3,350.31 \end{gathered}

The third investment is in a stock plan. The initial value of the investment is

P = 20% of $20,000 = $4,000

By the end of the first year, the stock plan increased by 8%, thus its value is:

M1 = $4000 * 1.2 = $4,800

By the end of the second year, the stock plan decreased by 4$, thus the value is:

M2 = $4,800 * 0.96 = $4,608

Finally, the stock plan increases by 6%, resulting in a final balance of:

M3 = $4,608 * 1.06 = $4,884.48

Finally, the last investment is in a savings account at 2.90% APR compounded annually for 3 years (not mentioned, but assumed).

P = $20,000 - $9,000- $3,000 - $4,000 = $4,000

i = 2.90 / 100 = 0.029

n = 3

Calculating:

\begin{gathered} M=\$4,000_{}(1+0.029)^3 \\ M=\$4,000\cdot1.089547389 \\ M=\$4,358.19 \end{gathered}

To summarize, the final balances for each type of investment at the end of the third year are:

Investment 1; $10,226.33

Investment 2: $3,350.31

Investment 3: $4,884.48

Investment 4: $4,358.19

Total balance: $22,819.32

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