Answer:
A) an upper quartile of 56 and a median of 50
Step-by-step explanation:
Hello!
The lower and upper ends "box" from the box and whiskers plot is determined by the first (Q₁) and third (Q₃) quartiles of the data set and the line inside the box marks the median or second quartile of the sample (Q₂)
The length of the whiskers of the diagram is determined by the minimum value (left whisker) and the maximum value (right whisker)
If you observe the graphic:
Q₁= 42
Q₂= 50 (Median)
Q₃= 56
Min= 38
Max= 66
The box is delimited by a lower quartile of 42 and an upper quartile of 56, with a median 50, the least observation is 38 and the highest observation is 66.
The correct choice is A.
I hope this helps!
Answer:
1.7 seconds
Step-by-step explanation:
we have

where
v is the initial velocity (in feet per second)
s represents the initial height (in feet)
In this problem we have
---> because the object is dropped

substitute

Remember that
When the chestnut hit the ground the value of H is equal to zero
so
For H=0

solve for t

therefore
The solution is t=1.7 sec
A. = logb (3/7)^1/2
= logb 3^1/2 - logb 7^1/2
= 1/2(logb 3 - logb 7)
b. = log 7x^3 + log y^5
= log 7 + log x^3 + log y^5
= log 7 + 3 log x + 5 log y
The least number of chaperones they will need is 5 chaperones because the greatest amount of students in a group is 4, so less chaperones. If each group has one chaperone that is 5 people per group total. 25 divided by 5 is 5.