Six times a number a number increased by three is negative twenty-one. Find the number and write the equation

Linear relationships are important to understand because they are common in the world around you. For example, all rates and rat

emmasim [6.3K]

Two other examples of linear relationships are changes of units and finding the total cost for buying a given item x times.

<h3>Other examples of linear relationships?</h3>

Two examples of linear relationships that are useful are:

Changes of units:

These ones are used to change between units that measure the same thing. For example, between kilometers and meters.

We know that:

1km = 1000m

So if we have a distance in kilometers x, the distance in meters y is given by:

y = 1000*x

This is a linear relationship.

Another example can be for costs, if we know that a single item costs a given quantity, let's say "a", then if we buy x of these items the total cost will be:

y = a*x

This is a linear relationship.

So linear relationships appear a lot in our life, and is really important to learn how to work with them.

If you want to learn more about linear relationships, you can read:

brainly.com/question/4025726

6 0

2 years ago

Determine whether each of the following are functions

Alja [10]

Answer:

xxc

Step-by-step explanation:

7 0

3 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

Can somebody help me please ?

Maurinko [17]

Answer:

ik that angle one would be the same as 3 and 2 and 4 are also the same so if u find an answer for 2 or 4 they would be the same but is there like a total of the entire thing

7 0

3 years ago

Read 2 more answers

The basic wage earned by a truck driver for a a 40 hour week is 560.00$ for overtime work the driver is paid one and a half hour

Lubov Fominskaja [6]

Answer:

$7

Step-by-step explanation:

overtime = 1/2 x daily hourly rate

daily hourly rate = amount earned in a week / total hours worked

560 /40 = 14

overtime = 14/2 = 7

6 0

2 years ago