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3 years ago

Find the slope and y-intercept. Pleasee

Mathematics

2 answers:

masha68 [24]3 years ago

6 0

Answer:

$slope=-2\\\\y-intercept=6$

Step-by-step explanation:

The slope is the change in the y-axis over the change in the x-axis. To find the slope, you need to move vertically from a given point. Once you are horizontal to another point, move over to it. Count the spaces between each:

Start at (0,6)
Move down to (0,4). Since you moved down two spaces, use the value -2 to define the rise
Move to the right to point (1,4). Since you moved to the right 1 space, use the value 1 to define the run.

Insert the values into $\frac{rise}{run}$ → $\frac{-2}{1}$

Simplify → $-2$

Therefore, the slope is -2.

The y-intercept is where $x=0$ and where the line crosses over the y-axis. The coordinate point where this happens is $(0,6)$ , so the y-intercept is 6.

:Done

tatyana61 [14]3 years ago

3 0

Answer:

slope: -2

y-intercept: 6

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

A two sample t-test is used to compared the means of two samples who were exposed to different scenarios and therefore, it helps us to see the difference between these two samples.

In this problem, the teacher is experimenting with a new computer-based instruction. To be able to use the two-sample t-test she would have to have two randomly assigned samples and expose each of these two samples to different scenarios (one would receive a traditional instruction and the other one would receive a computer instruction), and finally she would need to compare the scores in both samples.

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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)

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Step-by-step explanation:

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