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andrey2020 [161]
2 years ago
12

Use the rational zeroes theorem to write the

Mathematics
1 answer:
scoundrel [369]2 years ago
8 0

The y factors of the given quadratic equation is -1 and 40/22.

According to the statement

we have given that the equation and we have to find the factors of those equation.

So, The given equation is:

Y =x^2-23x^2 + 18x+40

where y is the factors of that equation.

So, we have to find the factors of the given quadratic equation.

x^2-23x^2 + 18x+40 = 0

Then after solving it become

-22x^2 + 18x+40 = 0

Then take negative sign common from it then

22x^2 - 18x - 40 = 0

Make the factors of this quadratic equation

22x^2 + 22x -40x - 40 = 0

Now take common values 22x and 40 from the equation then

22x (x + 1) -40 (x + 1) = 0

Then it become

(22x - 40) (x + 1) = 0

Then the values of x become from the equation is

x = -1, x= 40/22

The value of the factors y is -1 and 40/22.

So, The y factors of the given quadratic equation is -1 and 40/22.

Learn more about Quadratic equation here

brainly.com/question/24334139

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What is greater 0. 023 or 0. 3.
Veronika [31]

Answer:

The answer is 0.023<0.3

5 0
3 years ago
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

7 0
3 years ago
I need help with this question plz
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A) The estimated total expense is
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X+2(x+3(x+4(x+5)))=153
Allushta [10]

Answer:

x=1

Step-by-step explanation:

Here first open the inner parentheses and use the distributive property,

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3 0
3 years ago
Natalie buys organic almonds priced at $77 from the grocery store. How much did she pay the cashier, if she received $23 in chan
VARVARA [1.3K]

Answer:

$100

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You just add 77 and 23 which is 100. Therefore she gave the cashier 100 dollars.

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