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2 years ago

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A landscaping company charges a $12 fee to make a house call plus an hourly rate for labor. If the total bill for a job that tak

es 2.5 hours comes to $95.75, write an equation that can be used to solve for the hourly labor rate. Use r to represent the rate.

1 answer:

quester [9]2 years ago

5 0

Answer:

$y=41,8 . x + 12$

Step-by-step explanation:

This problem is solved using the linear function formula:

$y=a.x+b$

Where:

y= total bill

x= number of hours worked

b=fee

We have to find "a" in order to have the complete equation. We know that the constant fee (b) is 12$, and that if the hours worked are of 2.5 the total bill is $95.75, so we supplant on the equation and isolate "a".

$95.75=a. 2,5 + 12$

$95.75-12=a. 2,5$

$95.75-12=a. 2,5$

$83.75=a. 2,5$

$83.75 : 2=a$

$83.75 : 2=a$

$41,8=a$

So now we have the complete equation:

$y=41,8 . x + 12$

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610 people go on a coach trip each coach can hold 55 people how many coaches are needed

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Answer:

12

Step-by-step explanation:

610 divided by 55 = 11.09

but since u cant have 11.09, u need 12

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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

A parcel has a mass of 500g. Calculate its weight. (Assume the gravitational field strength is 10N/kg).

Elan Coil [88]

Answer:

The weight of the parcel is 5 N

Step-by-step explanation:

we know that

Weight can be calculated using the equation:

$W=mg$

where

W ----> weight (W) is measured in newtons (N)

m ----> mass (m) is measured in kilograms (kg)

g ----> gravitational field strength (g) is measured in newtons per kilogram (N/kg)

In this problem we have

$g=10\ N/kg$

$m=500\ g$

Remember that

$1\ kg=1,000\ g$

To convert g to kg

Divide by 1,000

so

$m=500\ g=500/1,000=0.5\ kg$

substitute the values in the equation

$W=(0.5)(10)=5\ N$

therefore

The weight of the parcel is 5 N

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2 years ago

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Nat2105 [25]

"4600" (and any subsequent words) was ignored because we limit queries to 32 words.

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2 years ago