Which is most likely the correlation coefficient for the set of data shown?

Mathematics

2 answers:

elena55 [62]2 years ago

7 0

Answer:

0.872

Step-by-step explanation:

The line is increasing. This means the correlation coefficient will be positive.

This is a fairly good fit; this means we want the larger of the two positive numbers, or 0.872.

mafiozo [28]2 years ago

6 0

That's a really good positive correlation pictured. The line fits the data very well. The slope of the line is positive, meaning as x increases so does y. I'd choose a big positive value.

Answer: 0.872

You might be interested in

Thomas is planning a trip to India. He is currently deciding which cities he would like to visit during his stay. The table belo

Oksana_A [137]

Swap Thiruvananthapuram for Jaipur and drop Bengaluru and Swap Mumbai for New Delhi, Thiruvananthapuram for Jaipur, and Bhubaneswar for Lucknow.this option is cheaper for thomas

4 0

2 years ago

Read 2 more answers

Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$