Answer: x=-12
Step 1: Simplify both sides of the equation.
4x+27=−33−x
4x+27=−33+−x
4x+27=−x−33
Step 2: Add x to both sides.
x+27+x=−x−33+x
5x+27=−33
Step 3: Subtract 27 from both sides.
5x+27−27=−33−27
5x=−60
Step 4: Divide both sides by 5.
x=−12
Answers:
Q: What is the most they could weigh together?
A: 0.74 kg
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Q: What is the least they could weigh together?
A: 0.62 kg
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Work Shown:
x = weight of first ball
y = weight of second ball
each ball has a weight range of 0.31 kg to 0.37 kg, so,
add straight down to get
which simplifies to
the two soccer balls have a weight range of 0.62 to 0.74, inclusive of both endpoints.
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Without using algebra, you basically just add the smallest the two weights could be (0.31) to itself to get 0.31+0.31 = 0.62 which represents the smallest the two weights combined can be. The same happens with the largest weight of 0.37 to get 0.37+0.37 = 0.74 as the max weight of both objects together.
Answer:
Null hypothesis: <em>μ₁</em> = <em>μ₂</em>.
Alternate hypothesis: <em>μ₁</em> > <em>μ₂</em>.
Step-by-step explanation:
The experimenter want to see whether laboratory rats would, on average, eat more when confined to a crowded cage or an uncrowded one.
(1)
The null hypothesis for this test is defined as:
<em>H</em>₀: The rats would, on average, not eat more when confined to a crowded cage or an uncrowded one, i.e. <em>μ₁</em> = <em>μ₂</em>.
(2)
The alternate hypothesis will be directional. This is because the experimenter wants to determine whether the rats eat more or not.
So the alternate hypothesis is:
<em>H</em>ₐ: The rats would, on average, eat more when confined to a crowded cage or an uncrowded one, i.e. <em>μ₁</em> > <em>μ₂</em>.
Answer:
The answer is B (From the cup to the hand)
Step-by-step explanation:
