All categories

1 year ago

What conclusion can be derived by comparing the central tendencies of the two data sets?

1 answer:

3 0

The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest

You might be interested in

Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a

Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field $F=-6i+5x^2j-5k$

We need to calculate the flux

Using formula of flux

$\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)$

Put the value into the formula

$\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz$

$\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz$

$\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}$

$\phi=682.6\ Wb$

Hence, The flux is 682.6 Wb.

7 0

1 year ago

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

Ann [662]

We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km

8 0

1 year ago

A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12

den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

$0.0192 kgm^{2}/s$

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

$L = \frac{1}{{12}}m{l^2}\omega$

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for $\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}$

5 0

2 years ago

Read 2 more answers

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ioda

Answer:

$Q_{week} = 458884.6\, BTU$

Explanation:

The weekly water consumption of Kimonoski is:

$m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$