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natita [175]
3 years ago
6

You walk an average of 3.2 miles per hour on the way to school and at an average of 4 miles per hour on the way home the total w

alking time is 18 minutes how far away is the school Round your answer to the nearest hundredth of a mile
Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0
The answer to this question is:
You walk an average of 3.2 miles per hour on the way to school and at an average of 4 miles per hour on the way home the total walking time is 18 minutes how far away is the school Round your answer to the nearest hundredth of a mile
<span>.53 mi 
</span><span>.133 hrs

Hope This Helped, Markadoodle
Your Welcome :)
</span>
guajiro [1.7K]3 years ago
5 0

Answer:

0.53 miles

Step-by-step explanation:

Given: You walk an average of 3.2 miles per hour on the way to school and at an average of 4 miles per hour on the way home the total walking time is 18 minutes.

To Find: how far away is the school.

Solution:

Average walking speed on the way to school =3.2\text {mph}

Average walking speed on the way to home =4\text {mph}

let total distance from home to school be = \text{D}

Distance from home to school = distance from school to home

time taken from home to school = \frac{\text{D}}{3.2} \text{hours}

time taken from school to home = \frac{\text{D}}{4} \text{hours}

Total time taken = 18 minutes = \frac{18}{60}\text{hours}

\frac{\text{D}}{3.2} \text{hours}+ \frac{\text{D}}{4} \text{hours}= \frac{18}{60}\text{hours}

\text{D}(\frac{1}{3.2}+\frac{1}{4})=\frac{18}{60}

\text{D}\frac{9}{16}=\frac{18}{60}

\text{D}=\frac{32}{60} \text{miles}

\text{D}=0.53 \text{miles}

The school is 0.53 \text{miles} from home

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It is often convenient to use "military" or "24-hour" time for calculations of time periods that extend over noon or midnight. For this purpose, 12 gets added to the hour number in the afternoon (pm). Of course, 39 minutes is 39/60 hours. That is found to be 0.65 hours by performing the division of 39 by 60.

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It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

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Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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