A) How many books were sold after 8 months?

Y= (x-9) (x+3)

juin [17]

Answer:

y = x² − 6x − 27

Step-by-step explanation:

To distribute, you can use something called FOIL. It stands for First, Outer, Inner, Last.

First, multiply the First term in each factor.

x · x = x²

Now multiply the Outer terms in each factor.

x · 3 = 3x

Next multiply the Inner terms in each factor.

-9 · x = -9x

Finally, multiply the Last terms in each factor.

-9 · 3 = -27

Add them all up:

x² + 3x − 9x − 27

x² − 6x − 27

4 0

3 years ago

choose two numbers from the list 3,4,6,12. Explain the difference between finding the greatest common factor and the least commo

zepelin [54]

3 & 6 - LCM = 3. 3's multiples are 3, 6, 9,12, 16, etc. 6's multiples are 6, 12, 18, etc. therefore, the least common multiple is 6.
GCF - 3's factors are 1, 3, and 6's factors 1, 2, 3. so the greatest common factor is 3.

6 0

3 years ago

The value of y varies directly with x and y=3 when x=-6. Find y when x=1

Iteru [2.4K]

Hi, ok all you need to do is use the direct variation formula which is= y/x=y/x so you will replace your numbers by each value which get us to = 3/-6=y/1 so you cross multiply which gives us 3=-6y you divide 3 by -6 and that is the value of y, thank you for reading hope it helps ;)

6 0

3 years ago

a local hamburger shop sold a combined total of 613 hamburgers and cheeseburger on Friday. there were 63 more cheeseburger sold

kondaur [170]

Answer:

275 hamburgers was sold on Friday

3 0

2 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$