PLEASE HELP AND NO LINKS OR YOU WILL BE REPORTED I need help with part B, I’m afraid I got this answer wrong because you are supposed choose the statements that are true and I’m afraid they are wrong, I chose the statements option 2, 3 and 5. The line plot in part A is correct so that can help you solve this as well but anyway the options are

The first is “The greatest length is 2 inches more than the least length.”

The second is “Less than half the lures are at least 3 inches long.”

The third is “The number of lures are greater than 2 1/2 inches long is less than the number of lures less than 2 1/2 inches

The fourth is “The greatest length is 2 inches.”

The fifth is “The number of lures is greater than 2 1/2 inches long is equal to the number of lures less than 2 1/2 inches.”

the answer is: the length is 5/6 yards.

give me a or a 5 star rating if this is correct!

step-by-step explanation:

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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.