50 POINTSS!!!!- The figure below shows the ideal pattern of movement of a herd of cattle, with the arrows showing the movement o
f the handler as he moves the herd. The arc the handler makes from the starting point to the return point should be a quarter of a circle:
Based on this theory, what distance will the handler move from the starting point to the return point if he creates an arc of a circle of radius 80 feet? (6 points)
125.6 feet
502.4 feet
83.73 feet
62.8 feet
Based on this theory, what distance will the handler move from the starting point to the return point if he creates an arc of a circle of radius 80 feet? (6 points)
125.6 feet
502.4 feet
83.73 feet
62.8 feet
2 answers:
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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.
Step-by-step explanation:
Salt in the tank is modelled by the Principle of Mass Conservation, which states:
(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)
Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:
By expanding the previous equation:
The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:
Since there is no accumulation within the tank, expression is simplified to this:
By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:
, where
.
The solution of this equation is:
The salt concentration after 8 minutes is:
The instantaneous amount of salt in the tank is:
Answer:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
Step-by-step explanation:
Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:
6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3
For this case we need to sort the data first on increasing way and we got:
1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4
For this case we have 9 values the median would be on the 5 position:
The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0
The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2
Then the interquartile rnage would be:
And
The lower limit on this case would be:
And our value is 1.2 is lower than the lower limit 1.7
Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below the first quartile.
False 1.2 is more than 1.5 IQR below the first quartile
C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the median.
False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median
D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the mean.
False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean
Chances of hitting bull's-eye is 
that =
1-:30= 0.03= 3%