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3 years ago

What can people do to stay safe before and during a storm with high winds and heavy rains?

1 answer:

7 0

Take shelter in a basement or any where underground. It is the best way so you don;t get effected if you were in a building abouve ground because windows can break and dangerous items may fly in from the wind. Hope this helps

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What is a good sentence for mass movement?

uranmaximum [27]

A mass movement of funds invested in the stock market could result in a stock market crash such as that of 1929.

8 0

3 years 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:

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

Read more about sampling distributions at:

brainly.com/question/10554762

3 0

3 years ago

Question 5 of 10

Masja [62]

A. I believe would be your answer

8 0

3 years ago

Since it was a school day denise thought it was peculiar

prisoha [69]

Answer:

what is the question?

Explanation:

no image, no real way to solve this problem

8 0

2 years ago

Suppose a friend of yours wants to buy a crystal candlestick. You remember seeing this item at both annie's attic and betty's br

Veronika [31]

The advice that will be given in the purchase of crystal candlestick is to go to

Betty's Breakables if the price is above $25.00.

In this scenario, we were told that Betty's Breakables offers 40% discount,

while Annie Attic offers $10 discount.

<h3>Calculation</h3>

Let's assume the price is $25.00, the discount at Betty's Breakables will be

= 40/100 × $25.00

= $10.

At Annie Attic, he was offered a $10 discount which is the same. However

there will be more gain on the discount if the price is above $25.

Let's assume the price is $30 , the discount will be $12 at Betty's Breakables

which is more than that at Annie Attic in which the discount remains at $10.

It is therefore cheaper to go to Betty's Breakable if the price of the

candlestick is above $25.

Read more about the Factors that influence pricing at brainly.com/question/17552787

5 0

2 years ago