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3 years ago

Please help me,I give up on this.i just don’t get it.

Mathematics

1 answer:

Stolb23 [73]3 years ago

8 0

Answer:

$y = -\frac{4}{5} x - \frac{14}{5}$
$y = \frac{5}{4} x -11$

Step-by-step explanation:

Given line ;

$y = \frac{5}{4} x -3\\\\m = slope\\m = \frac{5}{4}$

1. Perpendicular and passes through (4,-6)

$m_1 = \frac{5}{4} \\\\m_2 = \frac{-1}{m_1} \\\\m_2 = \frac{-1}{\frac{5}{4} } \\\\m_2 = -\frac{4}{5} \\\\(4,-6)=(x ,y)\\$

Substitute the values above into slope intercept form ; y=mx+b and solve for b.

$y =mx+b\\\\-6 = -\frac{4}{5} (4) + b\\\\-6 = -\frac{16}{5} + b\\\\-6 + \frac{16}{5} = b\\\\-\frac{14}{5} = b\\\\b = -\frac{14}{5}\\\\m = - \frac{4}{5}$

Substitute new values into ; y =mx+b.

$y = -\frac{4}{5} (x)- (\frac{14}{5} )\\\\y = -\frac{4}{5} x - \frac{14}{5}$

2.Parallel and passes through (4,-6)

$m = \frac{5}{4} \\\\(4,-6)=(x_1,y_1)$

Substitute values into point slope form and simplify

$y-y_1=m(x-x_1)\\\\y - (-6) = \frac{5}{4} (x -4)\\\\y+6 = \frac{5}{4} x -5\\\\y = \frac{5}{4} x -5-6\\\\y = \frac{5}{4} x -11$

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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

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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.

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