Based on the measurements of the pool and the path of uniform width, the area of the width and length of the path can be found to be 28.8 meters.

<h3>How to find the width and length?</h3>

First, find the area of the pool:

= Length x Width

= 10 x 26

= 260 meters ²

The area of the path can therefore be found to be:

= Total area of pool and path combined - Area of pool

= 1,092 - 260

= 832 meters²

Seeing as the path has uniform width, that means that the width is the same and the length so the width and length of the pool is:
= √area of the path

= √832

= 28.8

In conclusion, the width and length of the path are 28.8 meters.

Find out more on combined area at brainly.com/question/12966430

#SPJ1

8 0

1 year ago

A supplier of heavy construction equipment has found that new customers are normally obtained through customer requests for a sa

ziro4ka [17]

Answer:

The probability that it will take fewer than six customer contacts to clear the inventory is 0.8%.

Step-by-step explanation:

We have a probability of making an individual sale of p=0.15.

We have 4 units, so the probability of clearing the inventory with n clients can be calculated as:

$P=\dbinom{n}{4}p^4q^{n-4}=\dbinom{n}{4}0.15^4\cdot 0.85^{n-4}$

As we see in the equation, n has to be equal or big than 4.

In this problem we have to calculate the probability that less than 6 clients are needed to sell the 4 units.

This probability can be calculated adding the probability from n=4 to n=6:

$P=\sum_{n=4}^6P(n)=\sum_{n=4}^6 \dbinom{n}{4}0.15^4^\cdot 0.85^{n-4}\\\\\\P=0.15^4(\dfrac{4!}{4!0!}\cdot 0.85^{4-4}+\dfrac{5!}{4!1!}\cdot0.85^{5-4}+\dfrac{6!}{4!2!}0.85^{6-4})\\\\\\P=0.15^4(1\cdot0.85^0+5\cdot0.85^1+15\cdot0.85^2)\\\\\\P=0.00051(1+4.25+10.84)\\\\\\P=0.00051\cdot16.09\\\\\\P=0.008$