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3 years ago

The diagram represents 6x^2 - 7x + 2 with a factor of 2x - 1. What is the other factor of 6x^2 - 7x + 2?

Mathematics

2 answers:

Sever21 [200]3 years ago

8 0

Answer:

A. 3x - 2

Step-by-step explanation:

To find the factors on the left hand side of the table, factor out common factor from each row

GCF of 6x^2 and -3x

$6x^2= 3 \cdot 2 \cdot x \cdot x$

$-3x= -3 \cdot x$

GCF is 3x

GCF of -4x and 2

$-4x= -2 \cdot 2 \cdot x$

GCF is -2. Take out negative to make the first term positive.

So other factor is (3x-2)

Marizza181 [45]3 years ago

7 0

Answer:

Step-by-step explanation:

$\dfrac{6x^2}{2x}=3x,\ \dfrac{-3x}{-1}=3x\\\\\dfrac{-4x}{2x}=-2,\ \dfrac{2}{-1}=-2\\\\\underline{.\qquad|\ \ 2x\ |\ -1\ |}\\\underline{.\ 3x\ \ |\ 6x^2\ |-3x|}\\.\ -2|-4x|\ \ 2\ \ |$

$\text{Check:}\\\\(2x-1)(3x-2)=(2x)(3x)+(2x)(-2)+(-1)(3x)+(-1)(-2)\\\\=6x^2-4x-3x+2=6x^2-7x+2\qquad\text{CORRECT}$

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(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

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