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amid [387]
3 years ago
9

It took Everly 55 minutes to run a 10-kilometer race last weekend. If you know that 1 kilometer equals 0.621 mile, how many minu

tes did it take Everly to run 1 mile during the race? Round your answer to the nearest hundredth. (5 points) 3.41 minutes 5.50 minutes 6.21 minutes 8.86 minutes
Mathematics
1 answer:
AleksandrR [38]3 years ago
6 0
First convert to miles ----------- 0.621*10=6.21

Then divide the minutes by miles.

55/6.21 = 8.86 minutes
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0.4 repeating as a fraction for a mixed number
VashaNatasha [74]

Answer:

Step-by-step explanation:

1.Rewrite the decimal number as a fraction with 1 in the denominator

2.Multiply to remove 1 decimal places. Here, you multiply top and bottom by 101 = 10

3.Find the Greatest Common Factor (GCF) of 4 and 10, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 2,

so the answer is

0.4= 2/5

5 0
3 years ago
How much interest does $400 make in 5 years at 7% per annum?
ZanzabumX [31]

Answer:

$112

Step-by-step explanation:

7% for each year = $28

7% for four years = $28.4= $112

4 0
2 years ago
F(x) = 3x^4 is even or odd?
lara31 [8.8K]

Answer:

Step-by-step explanation:

it's even because f(-x)=f(x)

3 0
3 years ago
Read 2 more answers
A 6-sided die is rolled. what is the probability of rolling a number less than 4?
ohaa [14]

1/2

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All outcomes of a single die= 6

Events less than 4 =3

probability=3/6

=1/2

7 0
1 year ago
You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co
alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

P=2x+2y

On the other hand, since the available money to buy fence is D dollars,

\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}

Furthermore, the area of the enclosed land is given by

A=xy

Solving the second equation for x,

\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}

Substituting into the equation for the area,

\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}

To find the maximum possible area, solve A'(y)=0, as shown below

\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}

Therefore, the corresponding value of x is

\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}<h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2>\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}<h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0
11 months ago
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