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Liono4ka [1.6K]

2 years ago

12

The centers of two 15. 0-kilogram spheres are separated by 3. 00 meters. The magnitude of the gravitational force between the tw

o spheres is approximately.

1 answer:

ZanzabumX [31]2 years ago

7 0

The magnitude of the gravitational force between the two spheres is approximately; F = 1.67 x 10⁻⁹ N

<h3>Gravitational Force</h3>

We are given;

Mass 1; M = 15 kg

Mass 2; m = 15 kg

distance of separation; r = 3 m

Formula for gravitational force using newtons law of gravitation is;

F = GMm/r²

where;

M is mass 1

m is mass 2

G is gravitational constant = 6.67 × 10⁻¹¹ N.kg²/m²

r is distance of separation

Plugging in the relevant values, we have;

F = (6.67 × 10⁻¹¹ × 15 × 15)/3²

F = 1.67 x 10⁻⁹ N

Read more about Gravitational Force at; brainly.com/question/1602310

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Baker, schultz, and halstead argue that the united states could exploit significant strategic and economic gains if it invests in cleaner energy technologies and supports international efforts to reduce carbon emissions. True or false?

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1 year ago

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3 years ago

What is the plural forms of farm

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Answer:

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2 years ago

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul

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Sampling distribution involves the proportions of a data element in a given sample.

<em>The proportion of Good TV set is 0.67</em>
<em>The number of ways of selecting 5 from 6 TV sets is 6</em>
<em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

$^6C_5 = \frac{6}{1}$

$^6C_5 = 6$

Hence, there are 6 ways

<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

$^6C_4 = 15$

So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

$p = \frac{Good}{n}$

<u>(d) The sampling distribution</u>

In (a), we have:

$p = 0.67$ --- proportion of Good TV

The sampling error is calculated as follows:

$SE_n = |p - p_n|$

So, we have:

$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

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pshichka [43]

My best guess would be A. Add extra computer skills that you plan to learn.

And no need to be sorry, that's what we're here for. To help answer questions. :D

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3 years ago

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