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2 years ago

This is another trig question I need help with. Please explain how you do it to me I would appreciate it so much!!

Mathematics

1 answer:

Irina-Kira [14]2 years ago

3 0

Answer:

He stepped back 59.71 feet.

Step-by-step explanation:

Start by making 2 triangles. One has 68 degrees as the acute angle at the base of the triangle and the other has 41 degrees as the acute angle at the base of the triangle. The side opposite of said angle on both triangles will be 80, the height of the tree. You're solving for the bottom (horizontal) side for both triangles.

For the triangle with an acute base angle of 68, the formula to solve for the horizontal side is tan(68) = 80/x.

Get x by itself to get 80/tan(68) = x

x = 32.32.

For the next triangle, the equation is the same but the angle is changed to 41.

tan(41) = 80/x

80/tan(41) = x

x = 92.03

Finally, subtract the smaller distance from the larger:

92.03 - 32.32 = 59.71 steps.

You might be interested in

An employee earns $28.50 after 3 hours and $237.50 after 25 hours. What is the slope? What does the slope represent?

konstantin123 [22]

Answer:

You ad $28.50 and $237.50 and ad 25 and 3 so 266.00 divide into 28.

5 0

3 years ago

Pls help with this!!

ANEK [815]

A)
Let X be the food expenditure of a family.
Let the population mean of an expenditure of a family be.
Let the population standard deviation of an expenditure of a family be.
The proportion families spend more than $30 but less than $490 per month on food is,

7 0

2 years ago

2. The time between engine failures for a 2-1/2-ton truck used by the military is

OLEGan [10]

Answer:

A truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a random variable normally distributed (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The normal distribution is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

$\\ \mu = 6000$ miles.

$\\ \sigma = 800$ miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

We will use the concept of the standard normal distribution, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding z-score.
A z-score is a kind of standardized value which tells us the distance of a raw score from the mean in standard deviation units. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where x is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the standard normal table (available in Statistics books and on the Internet). We will use the cumulative standard normal table (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

$\\ P(x>5000)$ miles.

The z-score for x = 5000 miles is:

$\\ z = \frac{5000 - 6000}{800}$

$\\ z = \frac{-1000}{800}$

$\\ z = -1.25$

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations below the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

$\\ P(z$

Consulting a standard normal table available on the Internet, we have

$\\ P(z$

Then

$\\ P(z1.25)$

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

$\\ P(z>-1.25) = 1 - P(z$

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

$\\ P(z>-1.25) = 1 - 0.10565$

$\\ P(z>-1.25) = 0.89435$

In words, a truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of 12 trucks, and this is a problem of the sampling distribution of the means.

In this case, we have samples from a normally distributed data, then, the sample means are also normally distributed. Mathematically:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the samples means are normally distributed with the same mean of the population mean $\\ \mu$ , but with a standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ .

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z \sim N(0, 1)$

Then

The "average time-between-failures of 5000" is $\\ \overline{x} = 5000$ . In other words, this is the mean of the sample of the 12 trucks.

Thus

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{230.940148}$

$\\ z = -4.330126$

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

$\\ P(z$

The complement of P(z<-4.33) is:

$\\ P(z>-4.33) = 1 - P(z$ or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0

2 years ago

Another college precalculas question, please and thank you!

Ratling [72]

Answer:

B) (-∞,∞)

Step-by-step explanation:

The domain is the x-values, so it’s all real numbers in this graph.

I would use interval notation and choose B) (-∞,∞)

The range is the y-values, so it’s (2,-1), if that’s the next question.

5 0

2 years ago

Which of the following rational numbers is equal to 2 point 8 with bar over 8 ? twenty two over nine, twenty four over nine, twe

alina1380 [7]

Just to remove ambiguities, the bar over the expression means it's repeating itself to infinity.

$\bf 2.888\overline{8}\impliedby \textit{now say, let's make }x=2.888\overline{8}
\\\\\\
\textit{and multiply it by 10, to move over the 8 to the left}
\\\\\\
\begin{array}{llll}
10x&=&10\cdot 2.888\overline{8}\\
&&28.88\overline{8}\\
&&26+2.888\overline{8}\\
&&26+x
\end{array}\implies 10x=26+x
\\\\\\
9x=26\implies x=\cfrac{26}{9}$

notice, the idea being, you multiply it by 10 at some power, so that you move the "recurring decimal" to the other side of the point, and then split it with a digit and "x".

now, you can plug that in your calculator, to check what you get.

4 0

3 years ago

Read 2 more answers