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4 years ago

What is the median of the data set given below? 19,22,46,24,37,16,19,33

Mathematics

2 answers:

liq [111]4 years ago

6 0

Answer:

The median is 23

Step-by-step explanation:

katrin [286]4 years ago

6 0

Answer:

Step-by-step explanation:

Arrange the numbers in order from least to greatest

19,22,46,24,37,16,19,33

Least to greatest:

16, 19, 19, 22,24, 33, 37, 46

Then, find the number in the middle.

In this case, there isn't one number in the middle. There is 2: 22 and 24

To find the median, take the average of the 2 numbers.

Add them up and divide by 2

22+24=46

46/2=23

So, the median is 23

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A circular pool at a water treatment facility has a radius of 6 meters. About how long is the outer rim of the pool?

tia_tia [17]

I hope this helps you

Area=6^2pi=36pi

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7 0

3 years ago

Will mark brainliest! Please help!

ValentinkaMS [17]

Answer:

The dimensions are:

$3(2x+1)$ by $(7x+5)$

Step-by-step explanation:

The area of the rectangle is given as

$A=42x^2+51x+15$

The factored form of this quadratic trinomial gives the dimensions of the rectangle.

We factor 3 first to obtain;

$A=3(14x^2+17x+5)$

We split the middle term to get;

$A=3(14x^2+10x+7x+5)$

We factor within the parenthesis to get;

$A=3(2x(7x+5)+1(7x+5))$

We factor further to get;

$A=3(2x+1)(7x+5)$

The dimensions are:

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Then the perimeter will be

$2(6x+3+7x+5)=26x+16\:\:\:\boxed{\sqrt{} }$

4 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

What is the probability that you will roll a dice and get 3, then get an ace from a deck, then (without replacing the other card

pav-90 [236]

Answer:

que pasa parsero deja de esta mostrando el manda aquí vale

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3 years ago

Your sample size is 10 and your confidence is 90%. What is your t-score?

PilotLPTM [1.2K]

Answer:

if you want a t*-value for a 90% confidence interval when you have 9 degrees of freedom, go to the bottom of the table, find the column for 90%, and intersect it with the row for df = 9. This gives you a t*–value of 1.833 (rounded).

Step-by-step explanation:

Hope this helps!

6 0

3 years ago