What best describes the graph of y = -|x+1|

In 2009, Usain Bolt set the world record for sprinting 100m in approximately 9 3/5 seconds.

Hitman42 [59]

Answer:

10 5/12 is the answer, hope this helps

4 0

2 years ago

Show how to divide 9 pieces of licorice among 4 people

Ivanshal [37]

Answer:

1st question,<u> are you are able to cut the four pieces of licorice?</u> <u>if so cut them into 4 equal pieces and divide them among the 4 people</u>, making sure that each person gets an equal 9 pieces.

Step-by-step explanation:

9 pieces of licorice cut equally into 4 smaller bits then distributed throughout 4 people would play as this.

9x4=36

36/4 = 9

Each person would get 9 smaller pieces of licorice

6 0

2 years ago

How many miles will a person run during a 10 kilometer race?

faltersainse [42]

There are 1,61 kilometers in a mile or more correctly 1,60934 kilometers in a mile.
<span>But first, a short history.... Both kilometers and miles are units of measurement. They are used to measure distance. Miles are part of the imperial measurement system and kilometers are part of the metric measurement system. </span>
To calculate how many miles a person will run during a 10-kilometer race, you need to divide 10 kilometers by 1,61 miles (1,60934 rounded up to two digits).

Therefore your equation is as follows:

10÷1,61 or 10 000 meters ÷ 1 610 metres
1000 meters is equal to 1 kilometer.

The answer = 6, 21 miles.

Therefore, a person who runs 10 kilometers will run 6,21 miles

6 0

2 years ago

Read 2 more answers

Please help! i'm not good with algebra nor chemistry.

Snowcat [4.5K]

Here :2(6.02*10^23)(3.35*10^-23)

2*6.02*3.35

answer : 4.0334*10^1
hope this helped !

7 0

2 years ago

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

s344n2d4d5 [400]

The various answers to the question are:

To answer 90% of calls instantly, the organization needs four extension lines.
The average number of extension lines that will be busy is Four
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$$

Generally, the equation for is mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

$P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

What best describes the graph of y = -|x+1| - 5