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Aleonysh [2.5K]

2 years ago

13

Un guardia forestal divisa, desde un punto de observación A, un incendio en la dirección N27° 10’ E. Otro guardia, que se encuen

tra en el punto de observación B, a 6 millas al este de A, ve el mismo fuego en la dirección N52° 40’ W. Calcula, al décimo de milla más cercano la distancia de cada uno de los puntos de observación al incendio

1 answer:

Inga [223]2 years ago

8 0

Answer:

la distancia del punto de observación A al incendio = 5.5 millas

la distancia del punto de observación B al incendio = 3.8 millas

Step-by-step explanation:

La expresión esquemática de la pregunta se puede ver en la imagen adjunta a continuación :

Del siguiente diagrama;

Usando la sine regla:

$\dfrac{sin \ F}{f} = \dfrac{sin \ A }{a}$

$\dfrac{sIn \ 79}{6} = \dfrac{sin \ 63}{a}$

a sin 79 = 6 sin 63

$a =\dfrac{6 \times sin \ 63}{sin \ 79}$

$a =\dfrac{6 \times 0.8910}{0.9816}$

a = 5.446 millas

la distancia del punto de observación A al incendio = 5.5 millas (a la décima más cercana)

$\dfrac{sin \ F}{f} = \dfrac{sin \ B}{b}$

$\dfrac{sin \ 79}{6} = \dfrac{sin \ 38}{b}$

b sin 79 = 6 sin 38

$b = \dfrac{6 \times sin \ 38}{sin \ 79}$

$b = \dfrac{6 \times0.6157 }{0.9816}$

b = 3.7635 millas

la distancia del punto de observación B al incendio = 3.8 millas (a la décima más cercana)

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Answer:

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bonus15000

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Step-by-step explanation:

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1 year ago

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

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we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

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$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

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$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

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<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

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2 years ago

Connie keeps her sports card collection in a 54-page book. Each page has 9 slots for cards. Football cards take up the first 18

Ierofanga [76]

Answer:

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2 years ago

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What is the ratio of the afternoon to morning temperatures shown in the table? Temperatures Morning Afternoon 15 degrees. 30 deg

Nikitich [7]

Given:

Morning Temperature = 15 degrees, Afternoon Temperature = 30 degrees

Morning Temperature = 20 degrees, Afternoon Temperature = 40 degrees

Morning Temperature = 26 degrees, Afternoon Temperature = 52 degrees

To find:

The ratio of the afternoon to morning.

Solution:

Taking any one pair of morning and afternoon temperature, we can find the ratio.

Morning Temperature = 15 degrees

Afternoon Temperature = 30 degrees

$\text{Ratio}=\dfrac{\text{Afternoon Temperature}}{\text{Morning Temperature}}$

$\text{Ratio}=\dfrac{30}{15}$

$\text{Ratio}=\dfrac{2}{1}$

$\text{Ratio}=2:1$

Therefore, the correct option is C.

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2 years ago

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Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model

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Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

$\text{F}_(_n__1-1, n_2-1_)$

Here, $n_1$ = sample of 15 Claiborne models

$n_2$ = sample of 12 Klein models

So, the degrees of freedom = ( $n_1-1, n_2-1$ ) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The test statistics that will be used here is F-test which is given by;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $\text{F}_(_n__1-1, n_2-1_)$

where, $s_1^{2}$ = sample variance of the Claiborne models data = $\frac{\sum (X_i-\bar X)^{2} }{n_1-1}$ = 1.007

$s_2^{2}$ = sample variance of the Klein models data = $\frac{\sum (X_i-\bar X)^{2} }{n_2-1}$ = 2.653

So, the test statistics = $\frac{1.007}{2.653 } \times 1$ ~ $\text{F}_(_1_4,_1_1_)$

= 0.3796

Hence, the value of the test statistic is 0.3796.

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2 years ago