Drag each label to the correct location on the table. The owner of a sporting goods store is making a supply purchase for the coming month. Based on past experience, he has constructed a model
which shows that customers choose to buy soccer balls over baseball bats and tennis rackets 58% of the time.
The table below shows the results of three weeks of business with breakdowns for how many customers purchased soccer bals, baseball bats,
and tennis rackets.
Soccer Balls
85
110
64
Baseball Bats
B0
22
21
Tennis Rackets
B1
3
3
Week
2
Classify the results for each category as either consistent or inconsistent with the model.
Week 1 results
Week 2 results
Week 3 results
Total of all
3 weeks' results
Consistent with
Model
Inconsistent with
Model

c

step-by-step explanation:

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct

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.

apex answer is: true.