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3 years ago

Help, I need this soon!

Mathematics

1 answer:

jekas [21]3 years ago

3 0

Answer:

1a. x=2 1b. x= -1

Step-by-step explanation:

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An oval track is made by erecting semicircles on each end of a 44 m by 88 m rectangle. Find the length of the track and the area

Dmitry_Shevchenko [17]

Answer:

The length of the track is 247 m.

The area enclosed by the track is 5391 . 76 sq m.

Step-by-step explanation:

The length of the rectangle = 88 m

The width of the rectangle = 44 m

Now, as we know the semicircles are at the end of rectangular track.

So, the diameter of semicircles = Width of the rectangle

⇒ The diameter of the semi circle = 44 m

⇒Radius of the semi circle = 44/2 = 22 m

TOTAL LENGTH OF TRACK

= 2 ( Length of rectangle) + 2 (Circumference of 1 semicircle)

= 2 x (88 m) + 2 ( $\frac{2\pi r}{2}$ )

= 176 m + 2 (3.14)(22 m) = 176 + 71.08 = 247 m

Hence, the length of the track is 247 m

TOTAL AREA OF TRACK

= (Area of rectangle) + 2 (Area of 1 semicircle)

= (Length x Width) + 2 ( $\frac{\pi r^2}{2}$ )

= (44 m x 88 m) + (3.14)(22 m)(22 m) = (3872 + 1519.76) sq m

= 5391 . 76 sq m

Hence, the area enclosed by the track is 5391 . 76 sq m.

7 0

3 years ago

A rental truck company charges a flat fee of $30 to rent a truck in addition to $0.25 per miles how much would it cost to rent a

Sidana [21]

Answer:

$50

Step-by-step explanation:

x is distance and y is cost because the cost is dependent on the distance which makes the distance the independent variable. you can always use y=mx+b for these problems!!

7 0

2 years ago

A supplier of a certain car parts chain of stores wants to estimate the average length of time car owners plan to keep their car

____ [38]

Answer:

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{6.5}{\sqrt{25}} = 3.35$

The lower end of the interval is the sample mean subtracted by M. So it is 7.2 - 3.35 = 3.85 years

The upper end of the interval is the sample mean added to M. So it is 7.2 + 3.35 = 10.55 years

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

7 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

Help plsss i will mark you brainlest

kogti [31]

Answer:

x= 15

Step-by-step explanation:

6 0

3 years ago

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