How do you find the midpoint formula

Luden [163]

To find the equation of this line in slope-intercept form (y = mx + b, where m is its slope and b is its y-intercept), we naturally need the slope and the y-intercept. We can see that the line intersects the y-axis at the point (0, 4) so our y-intercept is 4, and the line rises 4 along the y-axis for every 2 it runs along the x-axis, so its slope is 4/2 = 2. With this in mind, we can write the line's equation as

y = 2x + 4

7 0

3 years ago

Can someone please answer this question please answer it correctly and please show work please help me I need it

I am Lyosha [343]

Answer:

the answer to this question is the number 8

5 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

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

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

he conditional relative frequency table was generated using data that compared the cost of one ticket for a performancehe method

ludmilkaskok [199]

The <u>correct answer</u> is:

0.14.

Explanation:

We are asked "Given that Lorenzo paid more than $30 for a ticket, what is the probability that he purchased the ticket at the box office?"

Using the conditional relative frequency table, we start at the column "More than $30". We then go to the row "Purchased at the Box Office." The value in this cell is 0.14; this is the probability.

7 0

4 years ago

Read 2 more answers

Question 5

Darina [25.2K]

Answer:

the answer is 12.24 .

Step-by-step explanation:

that answer is right is because when you add of those you get 530 , then you divide by how many numbers there are which is 7 , which equals to 75.7 . when you get that number ( 75.7) you subtract that number to 65,90,85,70,70,95, and 55 . when you get the total of all of those you add those . which is 10.7, 14.3 , 9.3 , 5.7 , 5.7 , 19.3 , and 20.7 . from adding those you get 85.7. you divide by 7 and get 12.24 . hope this helps :) .

8 0

2 years ago