Help with math 360×610÷409= what help

Slope -4 passing through the point (5,3).

bonufazy [111]

A negative slope means the line would be going downward, so you would go down first (instead of up as with a positive slope), then over right.

Therefore, you would go down 4 (or negative 4 considering you are going down) and right 1.

Rise over run still applies here. Rise over run is a fancy way of saying to put the number (or slope) in fraction form.

-4 as a fraction would be -4/1.

So start at (5,3), go down 4 and over 1, repeatedly and you got it.

**Just remember the rise over run can be confusing with negative slopes.**

5 0

3 years ago

Write an equation for the line parallel to the line 5/8x−7/6y =2 through the point (7,−1).

LiRa [457]

$y = - \frac{28}{48x - 15} - \frac{125}{321}$ is the answer.

Step-by-step explanation:

For this question,

• First, find the slope of the line parallel to the line 5/8x−7/6y=2 :-

$m = - \frac{28}{ - 15 + 48x}$

•Second, find the equation of the line,

$y = - \frac{28}{48x - 15} - \frac{125}{321}$

7 0

2 years ago

A) 120 is what percentage of 30

scZoUnD [109]

Answer:

a ) 400 percent

b ) n = 25

c) n= 36

Step-by-step explanation:

Is means equals, of means multiply and percents need to be changed to decimals.

a) 120 is what percentage of 30

120 = P * 30

Divide each side by 30

120/30 =P* 30/30

4 = P

Multiply by 100 to give the percentage

4 * 100

400 percent

b) 30 is 120% of what number

30 = 1.2 * n

Divide each side by 1.2

30/1.2 = 1.2n/1.2

25 = n

c) what number is 30%of 120

n = .3 * 120

n =36

3 0

2 years ago

Read 2 more answers

A golden rectangle is a rectangle whose length is approximately 1.6 times its width.the early greeks thought that a rectangle wi

Scrat [10]

Mike has 78 feet of fencing available for his garden, this is the perimeter (P) of the rectangle:
Perimeter: P=78 feet

The formula of Perimeter is:
P=2(W+L), where W is the width and L is the length, then:

P=78→2(W+L)=78
Dividing both sides of the equation by 2:
2(W+L)/2=78/2
W+L=39

If the shape is of a golden rectangle, we know:
L=1.6W
Replacing this above:
W+1.6W=39
Adding similar terms:
2.6W=39
Solving for W
2.6W/2.6=39/2.6
W=15 feet

L=1.6W=1.6(15)→L=24 feet

Answer: The dimensions of the garden are: Width=15 feet and Length=24 feet.

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 X = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, λt = 8 per hour.

(a)

For t = 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 t = 90 minutes = 1.5 hour, the value of λ, 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 t = 2.5 the value of λ, 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