Solve (4x^4-7-6x^2) (2x^2 4x^4 8)
1 answer:
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1, a.) The two specific conjectures are in the first image.
1, b.) The two general conjectures are in the second image.
2, a.) Two specific conjectures for this pattern are:
- The common difference between two consecutive terms is 3.
- And the given difference is A.P.
2, b.) From our observation two general conjecture is that the given sequence is an arithmetic sequence and the common difference is 3.
For finding its nth term we can use the formula: a(n) = a + (n-1) x d.
2, c.) A formula for the given pattern is 5 + (n-1)3, where n is the number of the term.
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!