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Yakvenalex [24]
2 years ago
8

A circle with a radius of 12 inches has an arc that measures 8 pi inches. Find the measure of the central angle determined by th

is arc.
Mathematics
1 answer:
ivanzaharov [21]2 years ago
3 0
We know that

[length of a circle]=2*pi*r
r=12 in
[length of a circle]=2*pi*12--------------> 24pi in

if 360° (full circle)--------------------> has a length of 24pi in
X-------------------------------------------> 8pi

X=8pi*360/24pi-----------> 120°

the answer is 120°

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sladkih [1.3K]
The error.....Kelvin just figured the 60% markdown cost...not the final price

the final price would be : 30.65 - 0.60(30.65) = 30.65 - 18.39 = 12.26
so 12.26 is the final price

or another way to do this is : if it is marked down 60%, then u r actually paying 40%......0.40(30.65) = 12.26
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The volume of an object that is 23 ft wide, 10 ft deep, and 8 ft height is:
monitta
The answer is 1840 because you have to do 10x8 then multiply that by 23 and get 1840
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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
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.

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2 years ago
Josh works for a construction company that uses 150 pounds of gravel for each driveway they pour. He can buy a dump truck load w
Pavlova-9 [17]

Answer:

Josh should by the dump truck with 30,000 pounds of gravel in it

Step-by-step explanation:

the easiest way to find this out is divide 30,000 and 2,500 by 150 and if the answer is a whole number, then you know that is the correct answer.

30,000/150 = 200

2,500/150 = 16.6666666666666

May I have brainliest please? :)

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2 years ago
after 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. what is the half-life of the sample?
coldgirl [10]
<span>(1/2)n = 0.015625 
n log 0.5 = log 0.015625 
n = log 0.5 / log 0.015625 
n = 6</span><span>24 days / 6 half-lives = 4.00 days (the length of the half-life)

</span>
6 0
2 years ago
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