The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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C. The upper quartile is shown by the end of the box.
Answer:
z=1
Step-by-step explanation:
3z+(-4)=-1
Add 4 to each side
3z+(-4) +4=-1+4
3z = 3
Divide each side by 3
3z/3 = 3/3
z=1
Answer:
8√π/π
Step-by-step explanation:
A = 64 = π·r²
r² = 64/π
= √64/√π
= 8/√π
= 8√π/π
Answer:
Mode = 81, Median = 81 and Range = 39.
Step-by-step explanation:
The mode is the most occurring value which is 81.
The median is also 81. There are a total of 19 numbers arranged in ascending order so the median is the 10th number.
The range is the highest - lowest number = 100 - 61 = 39.
