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3 years ago

8s+4(4s-3)=4(6s+4)-4

Mathematics

1 answer:

elena55 [62]3 years ago

5 0

Answer:

There is no solution.

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Find the area of each semicircle. Round to the nearest tenth. Use 3.14 for Pi.

pentagon [3]

Answer:

1. 69.4 / 2 = 34.7

2. 38.5 / 2 = 19.25

3. 201.1 / 2 = 100.55

I got these by caculating the area and dividing it by 2

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

−2x−(−5−4x)=−5(2x−7)

hoa [83]

Answer:

Step-by-step explanation:

- 2x - ( - 5 - 4x ) = - 5 ( 2x - 7 )

- 2x + 5 + 4x = - 10x + 35

- 2x + 4x + 10x = 35 - 5

12x = 30

<em>x = </em> $\frac{5}{2}$ <em> = 2.5</em>

Check the answer:

<em>L.H.S.</em> = - 2(2.5) - [ - 5 - 4(2.5) ] = <em>10</em>

<em>R.H.S.</em> = - 5 [ 2(2.5) - 7 ] = <em>10</em>

6 0

3 years ago

if a tv has a diagonal measurement of 55 and a height of 30, what is the tvs width? round to the nearest whole number

Sati [7]

Answer:

Step-by-step explanation:

The geometry can be modeled by a right triangle. The diagonal measure is the hypotenuse, and the height is one leg. The width is the other leg, and can be found using the Pythagorean theorem.

<h3>Pythagorean theorem</h3>

The relation between the leg lengths (a, b) and the hypotenuse (c) is ...

c² = a² +b²

Solving for b gives ...

b = √(c² -a²)

<h3>application</h3>

In this problem, we have c=55 and a=30. Then the width of the TV is ...

b = √(55² -30²) = √(3025 -900) = √2125

b ≈ 46.098

The width of the TV is about 46.

7 0

2 years ago

In matrix A = (aij) 3 x 3 , where aij= i+j then find the value of a33<br>

Tanya [424]

Answer: -9

Step-by-step explanation:

3 0

3 years ago

8s+4(4s-3)=4(6s+4)-4​

8s+4(4s-3)=4(6s+4)-4