Is (a-3)(2a^2 + 3a + 3) equivalent to 2(a^3 - 1.5a^2 - 3a - 4.5)
2 answers:
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Answer:
See Below.
Step-by-step explanation:
We are given the graph of <em>y</em> = f'(x) and we want to determine the characteristics of f(x).
Part A)
<em>f</em> is increasing whenever <em>f'</em> is positive and decreasing whenever <em>f'</em> is negative.
Hence, <em>f</em> is increasing for the interval:
And <em>f</em> is decreasing for the interval:
Part B)
<em>f</em> has a relative maximum at <em>x</em> = <em>c</em> if <em>f'</em> turns from positive to negative at <em>c</em> and a relative minimum if <em>f'</em> turns from negative to positive to negative at <em>c</em>.
<em>f'</em> turns from positive to negative at <em>x</em> = -2 and <em>x</em> = 1.
And <em>f'</em> turns from negative to positive at <em>x</em> = -1 and <em>x</em> = 3.
Hence, <em>f</em> has relative maximums at <em>x</em> = -2 and <em>x</em> = 1, and relative minimums at <em>x</em> = -1 and <em>x</em> = 3.
Part C)
<em>f</em> is concave up whenever <em>f''</em> is positive and concave down whenever <em>f''</em> is negative.
In other words, <em>f</em> is concave up whenever <em>f'</em> is increasing and concave down whenever <em>f'</em> is decreasing.
Hence, <em>f</em> is concave up for the interval (rounded to the nearest tenths):
And concave down for the interval:
Part D)
Points of inflections are where the concavity changes: that is, <em>f''</em> changes from either positive to negative or negative to positive.
In other words, <em>f </em>has an inflection point wherever <em>f'</em> has an extremum point.
<em>f'</em> has extrema at (approximately) <em>x</em> = -1.5, 0, and 2.2.
Hence, <em>f</em> has inflection points at <em>x</em> = -1.5, 0, and 2.2.
Answer:
1. Mean square B= 5.32
2. Mean square E= 16.067
3. F= 0.33
4. p-value: 0.28
Step-by-step explanation:
Hello!
You have the information of 3 groups of people.
Group 1
n₁= 20
X[bar]₁= 3.2
S₁²= 14.3
Group 2
n₂= 20
X[bar]₂= 4.2
S₂²= 17.2
Group 3
n₃= 20
X[bar]₃= 7.6
S₃²= 16.7
1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.
Df B= k-1 = 3-1= 2
Sum Square B is:
∑ni(Ÿi - Ÿ..)²
Ÿi= sample mean of sample i ∀ i= 1,2,3
Ÿ..= general mean is the mean that results of all the groups together.
General mean:
Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5
Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64
Mean square B= Sum Square B/Df B= 10.64/2= 5.32
2. The mean square error (MSE) is the estimation of the variance error (σ →
), you have to use the following formula:
Se²=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>
n₁+n₂+n₃-k
Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u> 915.8 </u> = 16.067
20+20+20-3 57
DfE= N-k = 60-3= 57
3. To calculate the value of the statistic you have to divide the MSB by MSE
4. P(F ≤ F) = P(F
≤ 0.33) = 0.28
I hope you have a SUPER day!