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3 years ago

I WILL GIVE BRAINLIEST Solve for x Answer must be simplified x - 6 ≤ 3

Mathematics

2 answers:

kobusy [5.1K]3 years ago

7 0

Answer:

x≤ 9

Step-by-step explanation:

Add 6 onto both side to isolate the variable.

6+3=9

Allisa [31]3 years ago

4 0

Answer:

x≤9

Step-by-step explanation:

x ≤6+3

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<img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B12x-7%3D0%20%5C%5C" id="TexFormula1" title="2x^{2} +12x-7=0 \\" alt="2x^{2}

m_a_m_a [10]

Answer:

-2. 46

- 9.54

Step-by-step explanation:

2x²+12x-7=0
2x²+12x+18= 25
2*(x²+6x+9)=25
2*(x+6)²= 5²
x+6= 5√2/2 ⇒ x= 5√2/2 -6 ≈ -2. 46
x+6= -5√2/2 ⇒ x= -5√2/2 -6 ≈ - 9.54

4 0

3 years ago

Identify the equation of a line in slope-intercept form that is perpendicular to y = -1/3 x + 2 and passes through (2, 1)

Airida [17]

Answer:

Step-by-step explanation:

The slope of a line perpendicular to another line is the opposite reciprocal of the slope of the other line, meaning if the slope of the other line is $-\frac{1}{3}$ , then the slope of the line perpendicular is $3$ .

Knowing this, we can start to construct the equation for the line:

$y = mx + b$

$y = 3x + b$

To solve for $b$ , we plug in the coordinates of the point that we know the line runs through, $(2, 1)$ :

$y = 3x + b$

$1 = 3(2) + b$

$1 = 6 + b$

$b = -5$

Therefore, the slope of the line that is perpendicular ot the line provided in the equation is the following:

$y = 3x - 5$

5 0

3 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:

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

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.

7 0

3 years ago