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2 years ago

Which of the following describes the graph of 2x + 4y < 16?

Mathematics

1 answer:

kherson [118]2 years ago

6 0

<h3>Answer: A) Dashed line, shaded below</h3>

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Explanation:

2x + 4y < 16 solves to y < -0.5x+4 when you isolate y. The inequality sign does not change direction because we divided both sides by a positive value (in this case, 4).

The graph of y < -0.5x+4 will be the same as the graph of 2x+4y < 16

To graph y < -0.5x+4, we graph y = -0.5x+4 which is a straight line that goes through the two points (0,4) and (2, 3). This is the boundary line of the inequality shaded region. The boundary line is a dashed line because we are not including points on the boundary that are part of the solution set. We only include these boundary points if the inequality sign has "or equal to".

We then shade below the dashed boundary line to indicate points below the boundary line. The shading is done downward due to the "less than" sign.

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Perhaps another method to find what direction we shade is we can try out a point like (0,0). The point cannot be on the boundary line.

Plug those coordinates into either equation. I'll pick the second equation

y < -0.5x+4

0 < -0.5*0+4

0 < 0+4

0 < 4

The last inequality is true, so the first inequality is also true when (x,y) = (0,0). Therefore, the point (0,0) is in the shaded region. The point (0,0) is below the boundary line y = -0.5x+4

So this is another way to see that the shaded region is below the boundary line.

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Answer:

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Step-by-step explanation:

the other nd 2 angles in the triangle both measures 30 degrees, the triangle angle sum theorem states that all the angles in a triangle add up to 180 degrees. Add the first two angles together and subtract it from 180. That's

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2 years ago

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 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:

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