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3 years ago

6

Hi I was wondering if someone know how to do this I did it but a long time ago and now the teacher wants me to do corrections to

improve my grade
The question that is asking is find the width of the rock formation AB

If someone could help me to start and give me steps or the process of how to do it I will really appreciated

1 answer:

Gennadij [26K]3 years ago

4 0

Answer:

I'll just help you start it

Step-by-step explanation:

Use similar triangles

CDE~ABE

48/84=54/AB

You might be interested in

How many terms of the series of - 3+0+3+6+9+...are needed to give a sum of 105?

dolphi86 [110]

Answer:

10

Step-by-step explanation:

Remember that the formula for the sum of an arithmetic series is:

$S=\frac{k}{2}(a+x_k)$

Where k is the number of terms, a is the initial term, and x_k is the last term of the series.

We essentially want to find k, the number of terms, given that the sum S is equal to 105. So, substitute 105 into our equation:

$105=\frac{k}{2}(a+x_k)$

To do so, we need to final term x_k. We don't know what it is yet, but that doesn't matter. All we need to do is to write it in terms of k. First, remember that the standard form for the explicit formula of an arithmetic sequence is:

$x_n=a+d(n-1)$

Where a is the first term, d is the common difference, and n is the nth term.

From our sequence, we can see that the first term is -3.

Also, we can determine that our common difference is +3, since each subsequent term is 3 <em>more</em> than the previous one. -3+3 is 0, 0+3 is 3, 3+3 is 6, and so on.

Therefore, our explicit formula is:

$x_n=-3+3(n-1)$

Therefore, our final term, x_k, will be if we substitute k for n. So, we can acquire the equation:

$x_k=-3+3(k-1)$

Now that we know what x_k is, we can substitute that into our original equation:

$105=\frac{k}{2}(a+x_k)$

Substitute the equation into x_k. Also, let's substitute -3 (our first term) for a. So:

$105=\frac{k}{2}(-3+(-3+3(k-1)))$

And now, all we have to do is to solve for k.

First, distribute the 3:

$105=\frac{k}{2}(-3+(-3+3k-3))$

Add within the parentheses:

$105=\frac{k}{2}(3k-9)$

Multiply both sides by 2. This removes the fraction on the right:

$210=k(3k-9)$

Distribute. We will get a quadratic:

$210=3k^2-9k$

So, let's solve for k. Let's divide everything by 3:

$70=k^2-3k$

Subtract 70 from both sides:

$0=k^2-3k-70$

Factor. We can use -10 and 7. So:

$0=(k-10)(k+7)$

Zero Product Property:

$k-10=0\text{ or } k+7=0$

Solve for k for each equation:

$k=10\text{ or } k=-7$

-7 doesn't make sense (we can't have -7 terms). Remove that solution. So, we are left with:

$k=10$

Therefore, the number of terms we have in our series for our sum to be 105 is 10.

And we're done!

4 0

3 years ago

You can represent the measures of an angle and its complement as x° and (90 − x)°. Similarly, you can represent the measures of

dimulka [17.4K]

Complement = 90
1)x+50=90
2)x=90-50
3)x=40

7 0

3 years ago

A news station would like to conduct an exit poll to determine the likelihood that a highly debated amendment will receive enoug

vladimir1956 [14]

Answer:

The expression is $n = (\frac{1.645*0.5}{0.03})^2$

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

What expression would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

We have to find n for which M = 0.03.

We have no prior estimate for the proportion, so we use $\pi = 0.5$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2$

$n = (\frac{1.645*0.5}{0.03})^2$

The expression is $n = (\frac{1.645*0.5}{0.03})^2$

3 0

3 years ago

If BC=CD=DE, what is th area of triangle ACD.

Lena [83]

I think 25 because 5×5 is 25

8 0

3 years ago

PLEAS HELP ASAP Fill in the missing probabilities on your paper and then answer the questions below. Make sure to type the ZERO

nikitadnepr [17]

This question is solved using probability concepts. We derive the probabilities from the tree given in the exercise, and with this, added to the use of conditional probability, we get the desired probabilities.

The probabilities are:

P(A) = 0.7, P(A and B) = 0.14, P(B) = 0.26, P(A or B) = 0.82, P(not B given A) = 0.8

Conditional probability:

In this problem, conditional probability concepts are used, and for this, we have that:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

P(A)

At the first node, we have that:

P(A) = 0.7

P(A and B):

From the first node, we have that $P(A) = 0.7$

From A to B, there is 0.2, which means that $P(B|A) = 0.2$

Thus

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

$0.2 = \frac{P(A \cap B)}{0.7}$

$P(A \cap B) = 0.7*0.2 = 0.14$

So

P(A and B) = 0.14

P(B):

P(B) = P(A and B) + P(not A and B).

P(not A) = 0.3, P(B|not A) = 0.4, then:

$P(not A and B) = 0.3*0.4 = 0.12$

P(B) = 0.14 + 0.12 = 0.26

P(A or B)

We have that:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We already have the three of them, so just replace:

$P(A \cup B) = 0.7 + 0.26 - 0.14 = 0.82$

Then

P(A or B) = 0.82

P(not B given A)

If A happens, either B happens, or it does not. That is:

P(B|A) + P(not B|A) = 1

Since P(B|A) = 0.2

P(not B|A) = 1 - 0.2 = 0.8

Then

P(not B given A) = 0.8

To take another look at conditional probability, you can check brainly.com/question/24161830

4 0

2 years ago