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3 years ago

How to factor a trinomial

Mathematics

2 answers:

Naddika [18.5K]3 years ago

8 0

You have to do the diamond method

Masja [62]3 years ago

6 0

You have to do the diamond method
Example
X^2-5x+6
a×c

b
6
-2 -3
-5

You might be interested in

Which statement is true about the extreme value of the given quadratic equation?

ICE Princess25 [194]

Answer:

Option B. The equation has a maximum value with a y-coordinate of -21.

Step-by-step explanation:

The correct quadratic equation is

$y=-3x^{2}+12x-33$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

Convert to vertex form

Factor -3

$y=-3(x^{2}-4x)-33$

Complete the square

$y=-3(x^{2}-4x+2^2)-33+12$

$y=-3(x^{2}-4x+4)-21$

Rewrite as perfect squares

$y=-3(x-2)^{2}-21$

The vertex is the point (2,-21)

therefore

The equation has a maximum value with a y-coordinate of -21

3 0

3 years ago

AB¯¯¯¯¯ and BC¯¯¯¯¯ are tangent to ⊙ O. Identify BC.

Volgvan

Answer:

Step-by-step explanation:

$2(3x-7)=4x+8\\x=11\\4(11)+8=52$

4 0

3 years ago

Of all the Harry Potter books purchased in a recent year, about 60% were purchased for readers 14yrs or older. If 12 Harry Potte

GaryK [48]

Using the binomial distribution, it is found that:

a) 0.9427 = 94.27% probability that at least 5 of them are 14 or older.

b) 0.1419 = 14.19% probability that exactly nine of the are 14 or older.

c) 0.0028 = 0.28% probability that less than 3 of them are 14 or older.

----------------------------

For each book, there are only two possible outcomes. Either it was purchased by readers 14yrs or older, or it was not. The probability of a book being purchased by a reader 14yrs or older in independent of any other book, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success on a single trial.

----------------------------

60% were purchased by readers 14yrs or older, thus $p = 0.6$
Sample of 12 books, thus $n = 12$

----------------------------

Item a:

The probability is:

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.6)^{0}.(0.4)^{12} \approx 0$

$P(X = 1) = C_{12,1}.(0.6)^{1}.(0.4)^{11} = 0.0003$

$P(X = 2) = C_{12,2}.(0.6)^{2}.(0.4)^{10} = 0.0025$

$P(X = 3) = C_{12,3}.(0.6)^{3}.(0.4)^{9} = 0.0125$

$P(X = 4) = C_{12,4}.(0.6)^{4}.(0.4)^{8} = 0.0420$

Then

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0 + 0.0003 + 0.0025 + 0.0125 + 0.0420 = 0.0573$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0573 = 0.9427$

0.9427 = 94.27% probability that at least 5 of them are 14 or older.

----------------------------

Item b:

This is P(X = 9), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.6)^{9}.(0.4)^{3} = 0.1419$

0.1419 = 14.19% probability that exactly nine of the are 14 or older.

----------------------------

Item c:

This is:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Using the probabilities from item a:

$P(X < 3) = 0 + 0.0003 + 0.0025 = 0.0028$

0.0028 = 0.28% probability that less than 3 of them are 14 or older.

A similar problem is given at brainly.com/question/15557838

6 0

3 years ago

Graph f(x) = 2x - 3 on a coordinate plane.

adoni [48]

Answer:

Create a table.

Plug values into the equation to complete the table.

Use the table to plot points on the graph.

Draw lines through the points on your graph.

The graph should look linear because there is an x that doesn't have a power greater than or less than 1.

5 0

2 years ago

A random sample of 20 married women showed that the mean time spent on housework by them was 29.8 hours a week with a standard d

DanielleElmas [232]

Answer:

95% confidence interval for the mean time spent on housework per week by all married women.

( 26.66 , 32.94)

Step-by-step explanation:

Step(i):-

Given random sample size 'n' = 20

Mean of the sample (x⁻ ) = 29.8 hours

Standard deviation of the sample (S) = 6.7

Given Margin of error = 3.14

Step(ii):-

95% confidence interval for the mean is determined by

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } , x^{-} +t_{0.05} \frac{S}{\sqrt{n} })$

We know that margin of error is determined by

$M.E = \frac{t_{0.05}XS.D }{\sqrt{n} } = 3.14$

Now 95% confidence interval for the mean time spent on housework per week by all married women.

$(29.8 - 3.14 , 29.8+3.14)$

( 26.66 , 32.94)

8 0

3 years ago