Slove it step by step

A property tax rate of 16.3 mills is equal to

Westkost [7]

It is approximately 1.000

5 0

3 years ago

4 question in one 65 points

jeka94

Q6.

The slope-intercept form: y = mx + b

m - slope

b - y-intercept

We have: slope m = 3, y-intercept (0, 4) → b= 4

<h3>Answer: y = 3x + 4</h3>

Q7.

2x + 4y = 4 |subtract 2x from both sides

4y = -2x + 4 |divide both sides by 4

y = -0.5x + 1

Only second graph has y-intercept = 1.

<h3>Answer: The second graph.</h3>

Q8.

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have

$(-16,\ 8)\to x_1=-16,\ y_1=8\\(4,\ -2)\to x_2=4,\ y_2=-2$

Substitute:

$m=\dfrac{-2-8}{4-(-16)}=\dfrac{-10}{20}=-\dfrac{1}{2}\\\\y-8=-\dfrac{1}{2}(x-(-16))\\\\y-8=-\dfrac{1}{2}(x+16)$

<h3>Answer: The first equation.</h3>

Q9.

It's a vertical line. The equation of a vertical line is x = <em>a</em>, where <em>a</em> is any real number.

<h3>Answer: x = -4</h3>

6 0

3 years ago

Read 2 more answers

Help please For the following function, construct a table.

AysviL [449]

Answer:

See the attachment below.

Step-by-step explanation:

Best Regards!

3 0

3 years ago

A daycare charges $115 per week for one child. Each additional child costs $65 for the week. If a family pays $505 for a week, h

valentinak56 [21]

Answer:

7 children

Step-by-step explanation:

$505-$115= $390

$390÷$65= 6

6+1= 7

4 0

2 years ago

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

$P(X\geq 6)=1-P(X$

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago