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3 years ago

Can you answer this? Take ur time I'm not in a hurry. Thanks!

Mathematics

1 answer:

il63 [147K]3 years ago

8 0

i believe the correct graph is the bottom left

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You are sitting at the edge of the Grand Canyon admiring the view when you notice a tour group at the bottom of the canyon. You

ololo11 [35]

Answer:

7949.13 feet

Step-by-step explanation:

We are given that,

Angle of depression from the spot to the group = 39°

Height of Grand Canyon = 5000 feet.

So, we will get the following figure of a right triangle.

<em>As, 'in a right triangle, the angles and the sides can be written in trigonometric form'.</em>

We get, $\sin A=\frac{perpendicular}{hypotenuse}$

i.e. $\sin 39=\frac{5000}{x}$

i.e. $x=\frac{5000}{\sin 39}$

i.e. $x=\frac{5000}{0.629}$

i.e. x = 7949.13 feet

Thus, the line of sight distance is 7949.13 feet.

5 0

3 years ago

I’m kinda confused. Can someone help.

horsena [70]

Answer:

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

Step-by-step explanation:

given MQR = 60, PQR = 75

MQP = 75 -60= 15 $QR = ((18(1+\cos 30))^{2}-(2\cos 15)^{2})^{\frac{1}{2}}$

WE KNOW THAT MQR +QRM +QMR = 180

MPQ = MQP AS ANGLE OPPOSITE TO EQUAL SIDES ARE EQUAL

THEREFORE QMR = 90 - 60 = 30

therefore PQM = 75 - 60 = 15

PM = PQ because M is the mid point

therefore PR = PM + $MQ\cos QMR$

PR = $18(1+\cos 30)$

$QR = (PR^{2}-PQ^{2}))^{\frac{1}{2}}$

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

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

There are 3 green marbles and 15 blue marbles in a bag. What is the ratio of green marbles to total number of marbles? Select al

Mandarinka [93]

3:18 is the answer hope this helps

8 0

2 years ago

Read 2 more answers

Which of the following is equal to 5 4/9? The answers are:

madam [21]

Answer:

5.4

Step-by-step explanation:

First work out 4/9 this is the same thing as 4 ÷ 9 which equals 0.44444

and then add the whole number 5 so,

5 + 0.44444 = 5.44444

3 0

3 years ago