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3241004551 [841]
3 years ago
11

ANSWER THIS QUICK BUT CORRECTLY FOR BRAINLIEST

Mathematics
1 answer:
nadya68 [22]3 years ago
6 0

Answer:

mean 7 and range is 22 median is 4.5

Step-by-step explanation:

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A bag contains tickets numbered 1 through 5. Use the probability distribution to determine the probability of drawing an even nu
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2 years ago
Question 8 of 10 What is sin 30º? 60 2 901 D A B. c.. () E. E )​
levacccp [35]

Answer:

A.  \displaystyle \frac{1}{2}

General Formulas and Concepts:

<u>Trigonometry</u>

  • [Right Triangles Only] SOHCAHTOA  
  • [Right Triangles Only] sinθ = opposite over hypotenuse

Step-by-step explanation:

<u>Step 1: Define</u>

sin 30°

<u>Step 2: Solve</u>

Since sin is opposite over hypotenuse, we need to find the length of the opposite leg of the angle and the length of the hypotenuse.

Given the triangle and it's values, the opposite leg to the 30° angle is 1 and the hypotenuse is 2.

Substituting it into sin, we have:

sin 30° = 1/2 = 0.5

∴ our answer is A.

3 0
2 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

7 0
3 years ago
the area of a rectangle of a barn is 200 square feet. the length is 10 feet longer than width . find the length and width of the
Elina [12.6K]

Area of rectangular barn = 200 (sqft)

Length= Width + 10

Length x Width = 200 ==> (Width+10)Width = 200

Width² +10Width -200 = 0===> width = 10  ===> length = 20

8 0
3 years ago
Read 2 more answers
Find the prime factorization: 49 i am stuck
ollegr [7]

Answer:

49 can be broken down into 7 * 7 so the prime factorization is simply 7².

8 0
3 years ago
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