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rewona [7]
3 years ago
7

Find the common ratlo for the following sequence.

Mathematics
1 answer:
worty [1.4K]3 years ago
3 0
The answer is D to the question
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Help me will give brainliest!!
faust18 [17]

Answer:

3/4 or 9/12

Step-by-step explanation:

Since there are 12 movies in total. The possible chance of picking a drama movie is a 1 out of 4 chance (3/12 = 1/4), the possible of picking a movie without a drama is a 3 out of 4 chance (9/12 = 3/4)

12/12 - 3/12 = 9/12 = 3/4

3 0
3 years ago
How to simplify 9m+4 (12-m)
Yuki888 [10]
5m+48 you will need to distribute 4 to the equation in the parenthesis first 9m+4(12-m) 9m+48-4m combine like terms 5m+48
7 0
3 years ago
Which fraction is halfway between 1/2 1 1/4
NemiM [27]

Answer:

7/8

its between 3/4 and 1

Step-by-step explanation:

think quarters

.50,.75, 1, 1.25

(1/2, 3/4, 1, 1 1/4)

good luck!!

3 0
2 years ago
The judiciary committee at a college is made up of three faculty members and four students. If ten faculty members and 25 studen
Andre45 [30]

Answer:

1,518,000 committees.

Step-by-step explanation:

We have been given that the  judiciary committee at a college is made up of three faculty members and four students. Ten faculty members and 25 students have been nominated for the committee.

We will use combinations for solve our given problem.

^nC_r=\frac{n!}{r!(n-r)!}, where,

n = Number of total items,

r = Items being chosen at a time.

^{10}C_3\cdot ^{25}C_4=\frac{10!}{3!(10-3)!}\cdot \frac{25!}{4!(25-4)!}

^{10}C_3\cdot ^{25}C_4=\frac{10!}{3!*7!}\cdot \frac{25!}{4!*21!}  

^{10}C_3\cdot ^{25}C_4=\frac{10*9*8*7!}{3*2*1*7!}\cdot \frac{25*24*23*22*21!}{4*3*2*1*21!}

^{10}C_3\cdot ^{25}C_4=10*3*4\cdot 25*23*22

^{10}C_3\cdot ^{25}C_4=120\cdot 12650

^{10}C_3\cdot ^{25}C_4=1,518,000

Therefore, 1,518,000 judiciary committees could be formed at this point.

7 0
3 years ago
Factor the following<br>(a²+b²)²-18(a²+b²)-88​
navik [9.2K]

Answer:

\rm\displaystyle(  {a}^{2}  +  {b}^{2} - 22)( {a}^{2}  +  {b}^{2}   +  4)

Step-by-step explanation:

we would like to factor the following:

\rm\displaystyle ( {a}^{2}  +  {b}^{2}  {)}^{2}  - 18( {a}^{2}  +  {b}^{2} ) - 88

let a²+b²=x

thus substitute:

\rm\displaystyle x {}^{2}  - 18x- 88

rewrite the middle term as 4x-22x:

\rm\displaystyle x^{2}  + 4x - 22x -  88

factor out x:

\rm\displaystyle x( x^{}  + 4)- 22x -  88

factor out -22:

\rm\displaystyle x( x^{}  + 4)- 22(x  +  4)

group:

\rm\displaystyle( x- 22)(x  +  4)

substitute back:

\rm\displaystyle(  {a}^{2}  +  {b}^{2} - 22)( {a}^{2}  +  {b}^{2}   +  4)

and we are done!

6 0
3 years ago
Read 2 more answers
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