A recent newspaper article claimed that more people read Magazine A than read Magazine B. To test the claim, a study was conducted by a publishing representative in which newsstand operators were selected at random and asked how many of each magazine were sold that day. The representative will conduct a hypothesis test to test whether the mean number of magazines of type A the operators sell, μAμA, is greater than the mean number of magazines of type B the operators sell, μBμB. What are the correct null and alternative hypotheses for the test? A. H0:μA−μB=0Ha:μA−μB>0H0:μA−μB=0Ha :μA−μB>0
B. H0:μA−μB<0Ha:μA−μB>0H0:μA−μB& lt;0Ha:μA−μB>0
C. H0:μA−μB=0Ha:μA−μB≠0H0:μA−μB=0Ha:μA −μB≠0
D. H0:x¯A−x¯B=0Ha:x¯A−x¯B>0H0:x¯A−x ¯B=0Ha:x¯A−x¯B>0
E. H0:μB−μA=0Ha:μB−μA>0H0:μB−μA=0Ha :μB−μA>0

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

sorry man i accidentally pressed yes my bad

step-by-step explanation:

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.