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AlekseyPX
2 years ago
13

in 2 h, kelly laid new floor tile over 1/5 of the room. when she was joined by annette, the rest of the work was completed in 3h

. how long would it have taken annette to do the entire job alone
Mathematics
1 answer:
jeyben [28]2 years ago
7 0

Answer: Annette will take 6 hours to do the entire job alone.

Step-by-step explanation:

Given: Time taken by Kelly to do \dfrac{1}{5} of job = 2 hours

i.e. Time for complete job done alone by Kelly =5\times2=10\ hours

Rest of work = 1-\dfrac{1}{5}=\dfrac{4}{5} of the job

\dfrac{4}{5}  of the complete job done by both Kelly and Annette in 3 hours

Time would be taken by then to do entire job together = 3\times\dfrac{5}{4}=3.75\ hours

Let t be the time taken by Annette to do job alone.

Then, as per situation

\dfrac{1}{3.75}=\dfrac{1}{10}+\dfrac{1}{t}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{1}{3.75}-\dfrac{1}{10}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{4}{15}-\dfrac{1}{10}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{1}{6}\\\\\Rightarrow\ t=6

hence, Annette will take 6 hours to do the entire job alone.

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Sasha has 3.20 in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could
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Answer:

10 quarters = $2.50

10 nickels = $0.50

that leaves $0.20 for other coins (dimes / pennies)

Step-by-step explanation:

First, suppose she has only quarters and nickels and no other coins.  Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this.  So she must have at least one other type of coin.

Assume she has only quarters, nickels, and dimes.  Then if D is the number of dimes, 5C +  25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32.  The smallest D can be is 2, leaving 3C = 30 and thus C = 10.  So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.

This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.

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According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
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a) \bar X=10.65

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And we can find the usual limits with:

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Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

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And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

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