Cracks in a rock from exposure to pressure are an example of

Which animal is native to the tundra?

Sergeu [11.5K]

Answer:

the polar bear

Explanation:

the polar bear is native to tundra

8 0

3 years ago

If you were as salesperson selling crm software, during the preapproach phase you would try to find out as much as possible abou

nydimaria [60]

A salesman marketing CRM software would ask inquiries regarding maintaining and preserving customer relationships.

<h3>Why is Customer Relationship Vital in Business Setting?</h3>

Customers' relationships are vital for company owners to establish long-term success. Developing a working connection with consumers is crucial for strategic and long-term success.

Customers feel more comfortable and engaged with a company when they have a solid connection based on trust and communication, which may lead to higher customer turnover and recurrent purchases.

Questions on customer wants can include;

"How do you keep in contact with potential customers and maintain connections with them?"

A salesman marketing CRM software would ask inquiries regarding maintaining and preserving customer relationships within the preapproach phase to gain a deep understanding of his customer's wants.

Learn more about customer relationships here:

brainly.com/question/25743859

8 0

1 year ago

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

lana [24]

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

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

Given

$n = 6$

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

(a) Proportion that are good

From the sample space, we have:

$Good = 4$

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

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

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

$p = 0.67$

(b) Ways to select 5 samples (without replacement)

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

(c) All possible sample space of 4

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.

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

(d) The sampling distribution

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:

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