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2 years ago

What is the y-intercept of the function,represented by the table of values below?

Mathematics

2 answers:

Hitman42 [59]2 years ago

8 0

Answer:

Step-by-step explanation:

So the y-intercept is not given by your table because there is no x that is listed as 0.

But don't fret; we can still find it.

Let's see if the function is linear by seeing if we have the same slope per two points in the table.

For the first pair ( the points (-2,16) and (1,4) ), x increased by 3 and the y decreased by 12 so the slope there is -12/3=-4.

Now looking at the next pair ( the points (1,4) and (2,0) ), x increased by 1 while y decreased by 4 so the slope is -4/1=-4.

So the function appears to be linear.

So the slope-intercept form of a line is y=mx+b where m is slope and b is y-intercept.

We already found the slope from earlier which is m=-4.

So the equation so far is y=-4x+b.

Now to find b, the y-intercept, we need to use a point (x,y) on the line along with y=-4x+b.

Let's see my favorite on the list of points is (2,0).

y=-4x+b with (x,y)=(2,0)

0=-4(2)+b

0=-8+b

8=b

So the y-intercept is 8.

Advocard [28]2 years ago

5 0

Answer:

Step-by-step explanation:

have a great day!:)

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2 years ago

Read 2 more answers

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution

oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

$P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!$

Then, we can calculate P(X ≤ 5) as:

$P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999$

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

$P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080$

4 0

1 year ago

The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin

Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.03

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0

2 years ago

given the points A(-3,-5) and B (5,0), find the coordinates of the point P on a directed line segment AB that partitions AB in t

Sonja [21]

$\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-3,-5)\qquad B(5,0)\qquad \qquad \stackrel{\textit{ratio from A to B}}{2:3} \\\\\\ \cfrac{A\underline{P}}{\underline{P} B} = \cfrac{2}{3}\implies \cfrac{A}{B} = \cfrac{2}{3}\implies 3A=2B\implies 3(-3,-5)=2(5,0)\\\\[-0.35em] \rule{31em}{0.25pt}\\\\ P=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] \rule{31em}{0.25pt}$

$\bf P=\left(\cfrac{(3\cdot -3)+(2\cdot 5)}{2+3}\quad ,\quad \cfrac{(3\cdot -5)+(2\cdot 0)}{2+3}\right) \\\\\\ P=\left( \cfrac{-9+10}{5}~~,~~\cfrac{-15+0}{5} \right)\implies P=\left(\frac{1}{5}~,~-3 \right)$