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astraxan [27]
3 years ago
11

What is 3c2+ 2c- 16?factoring​

Mathematics
2 answers:
Alex777 [14]3 years ago
6 0
<h3><u>Answer:</u></h3>

  • ( c - 2 ) ( 3c + 8 )

<h3><u>Solution</u><u>:</u></h3>

In the given question, we have to factorise the equation:

  • <u>3c² + 2c - 16</u>

Therefore, by using the middle term splitting method:

  • This method is used in equations which are in the form of ax² + bx + c. Here, we split the middle term into two terms which on multiplying gives ac and on adding or subtracting gives bx .

➙ 3c² + 2c -16

➙ 3c² - 6c + 8c - 16

➙ ( 3c² - 6c )+ (8c - 16 )

➙ 3c( c - 2 ) + 8 ( c - 2 )

➙ ( c - 2 ) ( 3c + 8 )

storchak [24]3 years ago
5 0

Factorisation of a quadratic polynomial of the type ax² + bx + c where (a ≠ 1) .

  • To factorise ax² + bx + c, we have to find two numbers whose sum is equal to the coefficient of x and product is equal to the coefficient of x² and constant term.

Consider the factorisation of 3c² + 2c – 16 .

We have to find two numbers whose sum is +2 and product (3 × 16) = 48 .

Obviously, the two numbers are 6 and 8 .

{ \qquad \sf { \dashrightarrow{   3c {}^{2}  + 2c  -  16  }}}

So, We can write it as,

{ \qquad \sf { \dashrightarrow{   3c {}^{2}  + 8c - 6c -  16  }}}

{ \qquad \sf { \dashrightarrow{   c (3c + 8) -2 (3c + 8) }}}

{ \qquad \sf { \dashrightarrow{  ( c -2 )(3c + 8) }}}

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4 years ago
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The proportion of brown M&amp;M's in a milk chocolate packet is approximately 14%. Suppose a package of M&amp;M's typically cont
marshall27 [118]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the scenario above :

A) State the random variable.

The random variable is the p opoertion of brown M&M's in a milk chocolate packet.

B.) Argue that this is a binomial experiment.

Each trial is independent for a total number of 52 trials with a set probability of success at 0.14

C) probability that 6 M&M's are brown:.

P(x) = nCx * p^x * (1-p)^(n-x)

p = 0.14 ; (1 - p) = 0.86 ; n = 52 ; x = 6

P(x = 6) = 52C6 × 0.14^6 × 0.86^46

= 20358520 × 0.00000752954 × 0.00097035078

= 0.1487

D) P(x =25)

P(x = 25) = 52C25 × 0.14^25 × 0.86^27

= 477551179875952 × 449.987958058*10^(-24) × 0.01703955245

= 0.00000000366

E) P(x = 52)

P(x = 52) = 52C52 × 0.14^52 × 0.86^0

= 1 × 3968.78758299*10^(-48) × 1

= 3968.78758299*10^(-48)

F) yes it would be unusual, because such probability is extremely low. However, if a huge or substantial number of trials such may occur

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Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o
faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

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S=365^n

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In point i:

That somebody mentions their birthday throughout the party

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In point ii:

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In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

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