HELP PLEASE I NEEED HELP!!!!

You are setting up a zip line in your yard. You map out your yard in a coordinate plane. An equation of the line representing th

Pachacha [2.7K]

Solving this problem needs the distance formula of point to a line.

The formula is:

distance = | a x + b y + c | / √ (a^2 + b^2)

So we are given the equation: y = 2 x + 4

Rewriting this would be: y – 2 x – 4 = 0 -> a = -2, b = 1, c = -4

We are also given the points:

(-4, 11) = (x, y)

Plugging it in the distance formula at points (x, y):

distance = | -2 * -4 + 1 * 11 + -4 | / √ [(- 2)^2 + (1)^2]

= 15 / √ (5)

= 6.7

So the tree is approximately 6.7 feet away from the zip line.

4 0

2 years ago

Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro

shepuryov [24]

The probability that both the chosen students are sophomores is 6/20 or 3/10 simplified.
the expresión that represents the probability that both students have chosen are sophomore is (6c1) (5c1) /(20c2)

5 0

2 years ago

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Which expression is the result of solving the equation ax - b = cy for 2? (For a + 0)

Masja [62]

I would say cy hopes this help

6 0

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

Reil [10]

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.

6 0

2 years ago

The number of pages that Allen, Kevin, Roneisha, and Ben can read in a day is shown below. Allen read 15% of his 56-page book. K

grandymaker [24]

It would be “Allen” since his percentage is far more grater then the rest, even Roneisha because even though her percentage is 1% greater, she has less pages, Ben has more pages but has 1% less.

Basically it’s like 2 bikers racing, 1 biker is faster but can’t go 5 miles, but the other is not that fast but can go 5+ miles.

Now who can go further, it would be the slow one.

3 0

1 year ago

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