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3 years ago

What are the mean and median of these numbers,7, 14, 15, 15, 14, 16, 11, 3

Mathematics

1 answer:

marissa [1.9K]3 years ago

8 0

Answer:

Mean: 11.875 or 11.9

Median: 14

Step-by-step explanation:

Mean = all numbers added up and then divided by the amount of numbers

Median = numbers ordered smallest to largest and then he middle number is your median, if there are two middle numbers then add those and divided by 2

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Need help on this question!

Anestetic [448]

Answer:

1 N 3

Step-by-step explanation:

8 0

3 years ago

Instructions: Given the following image of two parallel lines cut by a transversal, find the value of x.

Leya [2.2K]

Answer:

X=6

Step-by-step explanation:

10x-5=9x+1

x-5=1

x=6

8 0

2 years ago

Read 2 more answers

Write an explicit formula for an,the Nth term of the sequence 4,12,36

Rudik [331]

Answer:

$3^{n}+3^{n-1}$

Step-by-step explanation:

Since the sequence is geometric (it multiplies the previous number each time). The nth term sequence would be $x^n + yn$

Because the sequence multiplies by 3 each time, The base number in that equation would be $3^n +yn$ .

So $3^n$ of the equation would be:
$3, 9, 27$ .

Compare the 2 equations:
4, 12, 36

3, 9 ,27

The difference between the 2 equations is: 1, 3, 9 which itself is a geometric sequence so the nth term of this new equation is: $3^{n-1}$ .

Combine these 2 equations together and you get:

$3^{n}+3^{n-1}$ .

6 0

2 years ago

Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=

luda_lava [24]

Let $f(x) = x^3 - 2x - 2$ . Then differentiating, we get

$f'(x) = 3x^2 - 2$

We approximate $f(x)$ at $x_1=2$ with the tangent line,

$f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18$

The $x$ -intercept for this approximation will be our next approximation for the root,

$10x - 18 = 0 \implies x_2 = \dfrac95$

Repeat this process. Approximate $f(x)$ at $x_2 = \frac95$ .

$f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}$

Then

$\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}$

Once more. Approximate $f(x)$ at $x_3$ .

$f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}$

Then

$\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}$

Compare this to the actual root of $f(x)$ , which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0

2 years ago

There are 275 people in a movie theater. Of those, 150 are adults. What is the ratio of

Alexxandr [17]

Answer:

270 to 150

Step-by-step explanation:

3 0

3 years ago

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