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3 years ago

2

Mathematics

1 answer:

o-na [289]3 years ago

6 0

Answer:

ok so the answer would be h

Step-by-step explanation:

so basically the ratio is 4:2 -1:2 so purple is 2 and orange is 4. meaning purple is half of the orange. that ratio stays the same. if you were to multiply puplr by 60 you get 120. and if you multiply 4 by 60 you get 240. which 240 divided by 120 is 2 meaning h is the right answer choice.

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What is the common ratio of the sequence? -2, 6, -18, 54,...

marta [7]

Answer:1:3

Step-by-step explanation:

4 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

Determine the greatest common factor of the following numbers 8 and 12 NO LINKS PLEASE!!!!

jonny [76]

I think it would be 4.
8 breaks down to 4(2) and 12 breaks down to 4(3). Let me know if that helps!

3 0

3 years ago

Pls solve the simultaneous equation in the attachment.

siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> equation B

Multiply equation A by 10 both sides to remove the fractions

$5x-2y=10$ ----> equation C

isolate the variable y in equation B

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

$5x-2y=10$ ----> equation C

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

Multiply by 3 both sides

$15x-2x-48=30$

$15x-2x=48+30$

Combine like terms

$13x=78$

$x=6$

Find the value of y

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

The solution is the ordered pair (6,10)

Part b) we have

$xy=21$ ---> equation A

$x+5y=22$ ----> equation B

isolate the variable x in the equation B

$x=22-5y$ ----> equation C

substitute equation C in equation A

$(22-5y)y=21$

solve for y

$22y-5y^2=21$

$5y^2-22y+21=0$

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

Find the values of x

For y=1.4

$x=22-5(1.4)=15$

For y=3

$x=22-5(3)=7$

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0

4 years ago

Write an equation in Standard Form using the correct properties. Convert this equation to Slope-Intercept Form.

Lubov Fominskaja [6]

Answer:

I think what you are asking for is an example.

Step-by-step explanation:

Standard form equation

8x-4y = 20

1. move 8x to the other side

-4y=20-8x

2. divide both sides by -4

$\frac{-4y}{-4} = \frac{20-8x}{-4}$

3. Solve for this...

y=2x-5

7 0

3 years ago