Answer:
Jay's estimate is closer to the actual weight of the basketball
Step-by-step explanation:
The standard weight of an NBA basketball is 1.4 pounds.
Judging the accuracy of Jay's and Shawna's answer based on their percentage error when compared with the standard, we have that Jay's answer has a lower error as shown below
Error = (Standard value - Estimated value)/ Standard Value ×100
Error in Jays answer = 
× 100 = 64.28 %
Error in Shawna's answer = 
× 100 = 96.42%
From this, It is obvious that Jay's answer has a lower percentage error when compared to the actual value.
Answer:
K
Step-by-step explanation:
I THINK IT'S SHAPE IS SQUARE. IT HAS FOUR SIDES.
I THINK THE NAME IS TRUCK DIRVER DRUNKED LOL
TE-PAHADI GAMING
Answer:
Step-by-step explanation:
From the given information,
Suppose
X represents the Desktop computer
Y represents the DVD Player
Z represents the Two Cars
Given that:
n(X)=275
n(Y)=455
n(Z)=405
n(XUY)=145
n(YUZ)=195
n(XUZ)=110
n((XUYUZ))=265
n(X ∩ Y ∩ Z) = 1000-265
n(X ∩ Y ∩ Z) = 735
n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)
145 = 275+455 - n(X ∩ Y)
n(X ∩ Y) = 585
n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)
195 = 455+405-n(Y ∩ Z)
n(Y ∩ Z) = 665
n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)
110 = 275+405-n(X ∩ Z)
n(X ∩ Z) = 570
a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)
n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735
n(X ∪ Y ∪ Z) = 50
c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)
n(X ∪ Y ∪ C') = 145-50
n(X ∪ Y ∪ C') = 95
Answer:
What are you trying to find?
Step-by-step explanation:
Answer:
-8
Step-by-step explanation:
47-55=-8
