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1 year ago

3(x+y)=y, if (x,y) is a solution to the equation above and y cannot equal to zero what is the ratio x/y

1 answer:

6 0

*Given
3(x+y)=y
y is not equal to zero

*Solution

1. The given equation is 3(x+y) = y and we are tasked to find the ratio between x and y. Distributing 3 to the terms in the parenthesis,

3(x+y) = y
    3x + 3y = y

Transposing 3y to the right side OR subtracting 3y from both the left-hand side and the right-hand side of the equation would give

   3x = -2y

Dividing both sides of the equation by 3,

   x = (-2/3)y

Dividing both sides of the equation by y,

x/y = -2/3

Therefore, the ratio x/y has a value of -2/3 provided that y is not equal to zero.

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On a school bus trip, 15 out of 25 randomly selected students would like to stop for lunch. There are 55 students on the bus. Ba

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im pretty sure its 35/55

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2 years ago

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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Answer:

a) $\bar X=10.65$

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

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

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

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:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

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

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

Part 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.

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2 years ago

Of the 400 eighth-graders at pascal middle school, 117 take algebra, 109 take advanced computer, and 114 take industrial technol

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If 164 of the 400 eighth-graders take none of these courses, then 400-164=236 students take some courses.

Let x students take all three courses, then:

1. 70-x take only both algebra and advanced computer,

2. 34-x take only both algebra and industrial technology,

3. 29-x take only both advanced computer and industrial technology.

Now let's count how many student take only one course:

1. 117-(x+34-x+70-x)=13+x take algebra,

2. 109-(x+29-x+70-x)=10+x take advanced computer,

3. 114-(x+29-x+34-x)=51+x take industrial technology.

Now count all students that take some courses:

51+x+10+x+13+x+29-x+34-x+70-x+x=236,

x=236-51-10-13-29-34-70,

x=29.

Answer: all three courses take 29 students.

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