How to graph x>-2 on a number line
1 answer:
Hello!
To graph the solution x>-2 on a number, the first thing you need to do is to mark -2 on the number line using an open circle.
The reason you would use an open circle (like o instead of •) is because the symbols > and < represent a solution that is greater than or less than the number, but does not include it. In the case of x>-2, the solutions are all greater than than -2 but because the symbol does not state x can be -2 (if it were you would be using <span>≥</span> which states x can be any number greater than or equal to -2) all values of x are any numbers that are greater than -2 such as -1, 0, or 5.
Then, once you have plotted the open circle on -2, because all your solutions are greater than -2, draw a line on the number line (make it bold if you have to) going towards the right to show this.
To graph the solution x>-2 on a number, the first thing you need to do is to mark -2 on the number line using an open circle.
The reason you would use an open circle (like o instead of •) is because the symbols > and < represent a solution that is greater than or less than the number, but does not include it. In the case of x>-2, the solutions are all greater than than -2 but because the symbol does not state x can be -2 (if it were you would be using <span>≥</span> which states x can be any number greater than or equal to -2) all values of x are any numbers that are greater than -2 such as -1, 0, or 5.
Then, once you have plotted the open circle on -2, because all your solutions are greater than -2, draw a line on the number line (make it bold if you have to) going towards the right to show this.
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Answer:
Step-by-step explanation:
Hello!
You have two samples.
Sample 1 (time that takes one commute of worker 1)
n₁= 20
S₁²= 12.2
Sample 2 (time that takes one commute of worker 2)
n₂= 20
S₂²= 16.9
The hypothesis is that the first coworker is more consistent with his commute times, this means that the variability of his commute times is less than the variability of the commute times of worker 2. With this in mind, the hypothesis for a variance ratio test is:
H₀: σ₁² ≥ σ₂²
H₁:: σ₁² < σ₂²
α: 0.10
The statistic for this test is:
F= (S₁²/S₂²) * (σ₁²/σ₂²) ~ F
The test is one-tailed (left) with just one critical value:
(The F-table I use only has the values for high probabilities so I've used a conversion method to obtain the value for the small probability)
Your decision rule is:
If F ≤ 0.549, you reject the null hypothesis.
if F > 0.549, you support the null hypothesis.
The calculated value:
F= (S₁²/S₂²) * (σ₁²/σ₂²) = (12.2/16.9) * 1= 0.722
Since the calculated F-value is greater than the critical value, the decision is to not reject the null hypothesis. So you can say that the variability of the commute times of worker 1 is less than the variability of the commute times of worker 2.
I hope you have a SUPER day!