Answer:
800 pounds.
Step-by-step explanation:
We have been given that an object that weighs 24 pounds on Mercury weighs approximately 80 pounds on Earth. A rock weighs 240 pounds on Mercury. We are asked to find the weight of rock on Earth.
We will use proportions to solve our given problem as proportions state that the ratio between two proportional quantities is equal.

Upon substituting our given value, we will get:




Therefore, the weight of rock on Earth would be approximately 800 pounds.
Answer:
Choice A is correct
Step-by-step explanation:
The distribution for town A is symmetric, but the distribution for town B is negatively skewed. From the box plots it is clear that the tails of the box plot for town A are equal in length while for town B the left tail is longer implying a negatively skewed distribution.
Answer:
x=5,y=1 and z=-2
Step-by-step explanation:
We are given that system of equation
(I equation)
(II equation )
(III equation )
Equation II multiply by 3 then add with equation I
Then, we get
....(Equation IV)
Subtract equation II from equation III then we get
(equation V)
Adding equation IV and equation V then, we get

Substitute x=5 in equation V then, we get




Substitute x=5 and y=1 in equation then, we get




Hence, the solution for the given system of equation is given by
x=5,y=1 and z=-2
Your mode is 57 and your median is 2