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Anestetic [448]
2 years ago
6

What value of b makes the trinomial below a perfect square x^2-bx+100

Mathematics
1 answer:
Aloiza [94]2 years ago
6 0
1. Perfect square trinomials, are 2nd degree polynomials, of the form          a x^{2} +bx+c so that a \neq 0, b \neq 0, c \neq 0, which can be written as perfect squares.

2. For example (x+1) ^{2} = x^{2} +2x+1
 
(3x-1)^{2}= (3x)^{2}-2(3x)+(-1) ^{2}= 9x^{2}-6x+1    


3. Thus x^{2} +2x+1, 9x^{2}-6x+1 are perfect square trinomials.

4. x^{2} -bx+100= x^{2} -bx+ 10^{2}= (x+10)^{2}   or  (x-10)^{2}

5. In the first case -b=20, so b=-20. In the second case, -b=-20, so b=20.

6. b∈{-20, 20}
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Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random
Katyanochek1 [597]

Answer:

a) Null hypothesis: \mu_A =\mu_B =\mu C

Alternative hypothesis: \mu_i \neq \mu_j, i,j=A,B,C

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667  

And we have this property  

SST=SS_{between}+SS_{within}  

The degrees of freedom for the numerator on this case is given by df_{num}=df_{within}=k-1=3-1=2 where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-K=3*6-3=15.

And the total degrees of freedom would be df=N-1=3*6 -1 =15

The mean squares between groups are given by:

MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166

And the mean squares within are:

MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544

And the F statistic is given by:

F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326

And the p value is given by:

p_v= P(F_{2,15} >11.326) = 0.00105

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

b) (\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}

The degrees of freedom are given by:

df = n_B +n_C -2= 6+6-2=10

The confidence level is 99% so then \alpha=1-0.99=0.01 and \alpha/2 =0.005 and the critical value would be: t_{\alpha/2}=3.169

The confidence interval would be given by:

(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321

(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part a  

Null hypothesis: \mu_A =\mu_B =\mu C

Alternative hypothesis: \mu_i \neq \mu_j, i,j=A,B,C

If we assume that we have 3 groups and on each group from j=1,\dots,6 we have 6 individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667  

And we have this property  

SST=SS_{between}+SS_{within}  

The degrees of freedom for the numerator on this case is given by df_{num}=df_{within}=k-1=3-1=2 where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-K=3*6-3=15.

And the total degrees of freedom would be df=N-1=3*6 -1 =15

The mean squares between groups are given by:

MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166

And the mean squares within are:

MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544

And the F statistic is given by:

F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326

And the p value is given by:

p_v= P(F_{2,15} >11.326) = 0.00105

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

Part b

For this case the confidence interval for the difference woud be given by:

(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}

The degrees of freedom are given by:

df = n_B +n_C -2= 6+6-2=10

The confidence level is 99% so then \alpha=1-0.99=0.01 and \alpha/2 =0.005 and the critical value would be: t_{\alpha/2}=3.169

The confidence interval would be given by:

(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321

(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345

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