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3 years ago

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John has a rectangular-shaped field whose length is 62.5 yards and width is 45.3 yards. The area of the field is ________ square

yards. The perimeter of the field is _______ yards.

1 answer:

xxMikexx [17]3 years ago

3 0

2831.25 is the area and 215.6 is the perimeter

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While taking a ride in a hot-air balloon in Napa Valley, Francisco wonders how high he is. To find out, he chooses a landmark th

mihalych1998 [28]

The solution would be like this for this specific problem:

sin 54 degrees = Height/718.0624

Height = (sin 54 degrees)(718.0624) = 580.9 ft

So the height of the balloon is 580.9 ft. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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2 years ago

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Find cos(a) in the triangle.

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${\huge{\boxed{\mathcal{\green{ANSWER}}}}} \\ cos( \alpha ) = base \div hypotenuse \\ \\ base \: = \: 21 \\ \\ hypotenuse = 29 \\ \\ cos( \alpha ) = 21 \div 29 \\ \\ {\huge{\boxed{\mathfrak{\red{hope \: it \: helps \: }}}}}$

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2 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

2 years ago

Answer the questions from the picture <br> Step by step

andre [41]

Answer:

6 = 4 mins

Step-by-step explanation:

you have to take 200 and 50 and divide them

200 divided by 50 =4 and you have to tack mins to the end of it

id_k how many class rooms for number 5

and 7 i dk how many weeks there are therefore i cant solve them

7 0

1 year ago

Given the arc length of a circle is 127π centimeters and the subtended angle is 60˚. Find the Circumference of the circle.

Mumz [18]

The circumference of the circle is 762π cm.

<h3>What is length of arc of a circle?</h3>

The length of arc a circle can is determined with the radius and central angle of the arc.

The radius of the circle is calculated as follows;

$L = \pi r \times \frac{ \theta }{180} \\\\
\pi r \theta = 180 L\\\\
\pi r = \frac{180 L}{\theta}$

The circumference of the circle is calculated as follows;

$C = 2\pi r\\\\
C = 2(\pi r)\\\\
C = 2(\frac{180 \ L}{\theta } )\\\\
C = 2(\frac{180 \times 127\pi}{60 } )\\\\
C = 762 \pi \ cm$

Learn more about circumference of a circle here: brainly.com/question/9782777

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2 years ago