explain why there is never a single correct answer to a question that asks you to estimate some quantity

Someone pls help, its on trig ratios.

Mars2501 [29]

Answer:

$BC=12.8,\\AC=15.1$

Step-by-step explanation:

In right triangles only, the tangent of an angle is equal to its opposite side divided by its adjacent side.

Therefore, we have the following equation:

$\tan 58^{\circ}=\frac{BC}{8}$

Solving, we get:

$BC=8\tan 58^{\circ}\approx \boxed{12.8}$

Also in right triangles only, the cosine of an angle is equal to its adjacent side divided by the hypotenuse of the triangle.

$\cos 58^{\circ}=\frac{8}{AC},\\AC=\frac{8}{\cos 58^{\circ}},\\AC\approx \boxed{15.1}$

We can also use the Pythagorean theorem now that we've found BC. However, if you choose to do so, make sure you don't use a rounded value for BC, as that could cause a notable deviation in your final answer.

Using the Pythagorean theorem to verify our answers:

$8^2+(8\tan 58^{\circ})^2=\left(\frac{8}{\cos 58^{\circ}}\right)^2\:\checkmark$

6 0

3 years ago

The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it r

Alisiya [41]

What is the probability that it also rained that day answer 0.3

7 0

3 years ago

Read 2 more answers

I need to figure out if I'm doing this correctly

kogti [31]

To prove that triangles TRS and SUT are congruent we can follow these statements:

1.- SR is perpendicular to RT: Given

2.-TU is perpendicular to US: Given

3.-Angle STR is congruent with angle TSU: Given.

4.-Reflexive property over ST: ST is congruent with itself (ST = ST)

From here, we can see that both triangles TRS and SUT have one angle of 90 degrees, another angle that they both have, and also they share one side (ST) ,then:

5.- By the ASA postulate (angle side angle), triangles TRS and SUT are congruent

5 0

1 year ago

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

valkas [14]

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :

Null hypothesis: $\mu \geq 2$

Alternative hypothesis: $\mu < 2$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0

4 years ago

In triangle MNO shown below, segment NP is an altitude:

Effectus [21]

Answer: C. Definition of an Altitude

Step-by-step explanation:

Given: In triangle MNO shown below, segment NP is an altitude from the right angle.

Let ∠MNP=x

Then ∠PNO=90°-x

Therefore in triangle MNO,

∠MPN=∠NPO =90° [by definition of Altitude]

[Definition of altitude : A line which passes through a vertex of a triangle, and joins the opposite side forming right angles. ]

Now using angle sum property in ΔMNP

∠MNP+∠MPN+∠PMN=180°

⇒x+90°+∠PMN=180°

⇒∠PMN=180°-90°-x

⇒∠PMN=90°-x

Now, in ΔMNO and ΔPNO

∠PMN=∠PNO=90°-x

and ∠MPN=∠NPO =90° [by definition of altitude]

Therefore by AA similarity postulate, we have

ΔMNO ≈ ΔPNO

5 0

3 years ago