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zmey [24]
2 years ago
14

The sum of two integers is 54 and there difference is 16 what are the integers

Mathematics
1 answer:
Verizon [17]2 years ago
4 0
54/2 = 27 
27-8 = 19
27 + 8 = 35
19+35 = 54
35 - 19 = 16
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1 is the answer I think
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2 years ago
If your front lawn is 17.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snowflakes every minu
Alexus [3.1K]

Answer: 42.84 kg

Step-by-step explanation: Lawn is 17 ft x 20 ft.

Its area is S = 17 x 20 = 340 ft²

Each ft² of lawn accumulates 1050 snowflakes/min, 1ft² = 1050 s.f./min.

As 1 hour has 60 min, the lawn accumulates 63000/hour →

1050 x 60 = 63000

1ft² = 63000s.f./hour

This way, 340 ft² of lawn will accumulate 21,420,000 s.f/hour →

340 x 63000 = 21420000

As 1 snowflake has 2mg,  21,420,000 will have 42,840,000 mg.

As 1kg = 1,000,000 mg

42,840,000/1,000,000 = 42.84 kg

3 0
2 years ago
containers a and b hold 11,875 L of water together. conatainer b holds 2,391L more than container B holds. How many Liters of wa
Licemer1 [7]
Step 1 : 11875=2x+2391 Step 2: 11875-2391=2x Step 3: 9484=2x Step 4: 9484/2=x Step 5: x=4742
4 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 hypotenuse of right triangle ABC, line segment AC, measures 13 cm. The length of line segment BC is 5 cm.
slavikrds [6]
Cos ( ∠C ) = 5/13 = 0.38462
m ∠C = cos^(-1) 0.38642 = 67.4°
m ∠A = 90° - 67.4° = 22.6°
m ∠C - m ∠A = 67.4° - 22.6° = 44.8°
Answer:
B ) 44.8° 
7 0
2 years ago
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