Olympic athletes compete in a 3000 meter event, but not in two miles event. Which race is longer 3000 meters or 2.00 miles? (Given: 1 mi-1.61 km)

1. 3000 meters

2. 2.00 miles

1. 3000 meters

2. 2.00 miles

sondestiny120g
2022-08-12
Answered

1. 3000 meters

2. 2.00 miles

You can still ask an expert for help

asked 2022-04-05

Find the temperature at which the Celsius measurement and the Fahrenheit measurement are the same number.

asked 2022-04-02

Subjects were classified according to which of three groups they were assigned. Group A received lots of praise. Group B received moderate praise. Group C received no praise for correct answers to math problems. Following the manipulation, all subjects completed a posttest measure of mathematical ability. Higher scores indicate greater mathematical ability.

Does praise influence performance?

$\begin{array}{|ccc|}\hline \text{Group A}& \text{Group B}& \text{Group C}\\ 7& 4& 3\\ 6& 6& 2\\ 5& 4& 1\\ 8& 7& 3\\ 3& 5& 4\\ 7& 7& 1\\ \hline\end{array}$

Identify the IV and the scale of measurement;

Identify the DV and the scale of measurement and for the IV – identify the number of levels;

Identify null and alternative hypotheses, are they directional or non-directional?

Assume that the distributions of the populations are approximately normal.

Does praise influence performance?

Identify the IV and the scale of measurement;

Identify the DV and the scale of measurement and for the IV – identify the number of levels;

Identify null and alternative hypotheses, are they directional or non-directional?

Assume that the distributions of the populations are approximately normal.

asked 2022-06-04

I would like to run a Kalman filter over a set of measurements which may (will) be correlated. Essentially, each new measurement contains (say) 90% of the same information from the previous measurement. The measurements are made by integrating over time, and subsequent measurements are taken by overlapping integrations.

For various reasons, I can't necessarily decimate the measurements so that they are independent of each other, but if I apply each measurement at its full weighting, the filter will become over confident and its covariance will shrink too quickly.

My first thought is that there must be someway to increase the covariance from each measurement so that it doesn't overly affect the filter state, but I would like a rigorously correct solution, and I welcome any pointers.

A least squares approach which allowed expressing the correlated measurements would also be welcome, but I kind of prefer the Kalman filter because it gives me a running covariance as well as the filtered result.

Thank you in advance for any pointers! Is there a better stack exchange for this question?

For various reasons, I can't necessarily decimate the measurements so that they are independent of each other, but if I apply each measurement at its full weighting, the filter will become over confident and its covariance will shrink too quickly.

My first thought is that there must be someway to increase the covariance from each measurement so that it doesn't overly affect the filter state, but I would like a rigorously correct solution, and I welcome any pointers.

A least squares approach which allowed expressing the correlated measurements would also be welcome, but I kind of prefer the Kalman filter because it gives me a running covariance as well as the filtered result.

Thank you in advance for any pointers! Is there a better stack exchange for this question?

asked 2022-07-07

It appears that these notations are equivalent when referring to the measure with which a function $f(x)$ is integrated with respect to. It seems to me that the expression $\int fd{P}_{X}$ is very clear once some measure theory is learned. In what contexts are the other notations useful or necessary?

asked 2022-07-03

Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?

Is there any way to find it efficiently? I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D

Thank you.

Is there any way to find it efficiently? I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D

Thank you.

asked 2022-08-04

Approximate the rational number 32(287) to 1 significant figures.

asked 2022-06-21

Given ${f}_{k}\in {L}^{p}(\mathrm{\Omega})$, I have to prove that ${\Vert \sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}\u2a7d\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{\Vert {f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}$, where $\Vert f\Vert \doteq {\left({\displaystyle {\int}_{\mathrm{\Omega}}{\left|f\right|}^{p}}\right)}^{1/p}$ denotes the p-norm, with $1\u2a7dp<\mathrm{\infty}$. Since I've already proved the original Minkowski inequality, a simple induction estabilishes the finite case, i.e., ${\Vert \sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{m}{f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}\u2a7d\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{m}{\Vert {f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}(\ast )$, $\mathbb{N}\ni m<\mathrm{\infty}$.

So, what I've thought is, since ${L}^{p}(\mathrm{\Omega})$ is a Banach space, to suppose that $\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{\Vert {f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}<\mathrm{\infty}$, which would give us that $\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{f}_{n}$ converges in ${L}^{p}(\mathrm{\Omega})$, by completeness. Hence, the desired inequality follow by making $m\u27f6\mathrm{\infty}$ in $(\ast )$, but I'm not completely sure about this.

Any help will be appreciated.

So, what I've thought is, since ${L}^{p}(\mathrm{\Omega})$ is a Banach space, to suppose that $\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{\Vert {f}_{n}\Vert}_{{L}^{p}(\mathrm{\Omega})}<\mathrm{\infty}$, which would give us that $\sum _{n\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{\mathrm{\infty}}{f}_{n}$ converges in ${L}^{p}(\mathrm{\Omega})$, by completeness. Hence, the desired inequality follow by making $m\u27f6\mathrm{\infty}$ in $(\ast )$, but I'm not completely sure about this.

Any help will be appreciated.