What type of polygon is formed by the points (5,3), (3,-4), and (8,-4)

1. An observer 80 ft above the surface of the water measures an angle of depression of 0.7o to a distant ship. How many miles is

dybincka [34]

Answer:

1. The distance of the ship from the base of the lighthouse is approximately 1.24 miles

2. a)The horizontal distance the plane must start descending is approximately 190.81 km

b) The angle the plane's path will make with the horizontal is approximately 18.835°

3. The depth of the submarine is approximately 107.51 m

Step-by-step explanation:

The

1. From the question, we have;

The height of the observer above the water = 80 ft.

The angle of depression of the ship from the observer, θ = 0.7°

Let the position of the observer be 'O', let the location of the ship be 'S', let the point directly above the ship at the level of the observer be 'H', we have;

$tan(\theta) = \dfrac{Opposite \ leg \ length}{Adjacent \ leg \ length} = \dfrac{HS}{OH}$

$The \ horizontal \ distance \ of \ the \ ship, OH = \dfrac{HS}{tan(\theta) }$

HS = The height of the observer = 80 ft.

Therefore, we get;

$The \ horizontal \ distance \ of \ the \ ship, OH = \dfrac{80 \, ft.}{tan(0.7^{\circ}) } \approx 6,547.763 \ ft.$

The distance of the ship from the base of the lighthouse ≈ 6,547.763 ft. ≈ 1.24 miles

2. The elevation of the plane, h = 10 km

The angle of the planes path with the ground, θ = 3°

Similar to question (1) above, the horizontal distance the plane must start descending, d = t/(tan(θ))

∴ d = 10 km/(tan(3°)) ≈ 190.81 km

The horizontal distance the plane must start descending, d = 190.81 km

b) If the pilot start descending 300 km from the airport, the angle the plane's path will make with the horizontal, θ, will be given as follows;

From trigonometry, we have;

$tan(\theta) = \dfrac{Opposite \ leg \ length}{Adjacent \ leg \ length}$

Where the opposite leg length = The elevation of the plane = 10 km

The adjacent leg length = The horizontal distance from the airport = 300 km

$\therefore tan(\theta) = \dfrac{10 \, km}{300 \, km} = \dfrac{1}{3}$

$\theta = arctan\left(\dfrac{1}{3} \right ) \approx 18.835^{\circ}$

The angle the plane's path will make with the horizontal, θ ≈ 18.835°

3. The angle at which the submarine makes the deep dive, θ = 21°

The distance the submarine travels along the inclined downward path, R = 300 m

By trigonometric ratios, we have;

The depth, of the submarine, 'd' is given as follows;

$si(\theta)= \dfrac{Opposite \ leg \ length}{Hypotenuse \ length} = \dfrac{d}{R}$

∴ d = R × sin(θ)

d = 300 m × sin(21°) ≈ 107.51 m

The depth of the submarine ≈ 107.51 m

7 0

3 years ago

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th

Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value for pretest , y = test value for posttest

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90

The system of hypothesis for this case are:

Null hypothesis: $\mu_D \geq 0$

Alternative hypothesis: $\mu_D < 0$

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$

The next step is calculate the degrees of freedom given by:

$df=n-1=6-1=5$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(5)}$

The p value is lower than the significance level assumed $\alpha=0.05$ , so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.

3 0

3 years ago

Determine whether the 2 shapes are similar.

vichka [17]

Answer:

no

Step-by-step explanation:

cant explain jus nonono

8 0

3 years ago

26 x 69. Estimate <br><br><br>

vovikov84 [41]

Answer:

1750

Step-by-step explanation:

Round 26 to 25, round 69 to 70.

25 x 70= 1750

5 0

3 years ago

A new car is purchased for 17600 dollars. The value of the car depreciates at 11.75% per year. What will the value of the car be

Umnica [9.8K]

Answer:

Step-by-step explanation:

For compound interest, we use the formula A = P(1 + r/n)^{nt}, where A is the amount, P is the principle, r is the rate, n is the number of times compounded per period of time, and t is the time.

We can alter this for a decrease. Instead of adding r/n to 1, we can substract it:

A = 17,600 * (1 - 0.01175 / 1)^{1 * 11} = 15,454.24

After 11 years, the car will be worth $15,454.24

5 0

3 years ago