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Serjik [45]
10 months ago
12

What is the solution of the system of equations? 6x-2y = 4 -9x+3y = 12

Mathematics
1 answer:
katrin2010 [14]10 months ago
3 0

We will have the following:

First, we solve both equations for y, that is:

6x-2y=4\Rightarrow-2y=-6x+4\Rightarrow y=3x-2

&

-9x+3y=12\Rightarrow3y=9x+12\Rightarrow y=3x+4

From this we can see that they are parallel lines, that do not share any common points, so there is no solution for the system.

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Find the 64th term in the number sequence<br> 4,9,14,19
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319 is the term

Step-by-step explanation:

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        15 - ( -10x ) - 20 / ( 12 + ( -8 ) )

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What is 17. 5% of 360
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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature
Westkost [7]

Answer:

(a) The value of <em>k</em> is \frac{1}{15}.

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

(c) The probability that between two and four forms (inclusive) are required is 0.60.

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

Step-by-step explanation:

The random variable <em>Y</em> 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 <em>k</em>.

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

Thus, the value of <em>k</em> 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:

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

<u>Check condition 1:</u>

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.

<u>Check condition 2:</u>

\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 <em>y</em>.

7 0
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