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2 years ago

Factor the following trinomial completely 3x^2+15x+18

1 answer:

3 0

This is an easy problem

The first ring to do is divide all of the terms by 3

You get x^2+5x+6

Then you factor it normally
(x+2)(x+3)

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Find the Value of y.<br>_______________

nika2105 [10]

$\huge{ \mathfrak{ \underline{ Answer} \: \: ✓ }}$

The given figure is a parallelogram, therefore sum of adjacent angles measures 180°

So,

$y + 70 = 180$

$y = 110 \degree$

__________________________

$\mathrm{ \#TeeNForeveR}$

4 0

2 years ago

Karina wants to fill the base so that it is 85% full. What is the volume of the base that she is filling? (Find 85% of the volum

Mazyrski [523]

Question:

Karina has recently begun kickboxing and purchased training equipment as modeled in the diagram below. The cylindrical base is 15 inches tall with a diameter of 22 inches.

Karina wants to fill the base so that it is 85% full. What is the volume of the base that she is filling? (Find 85% of the volume).

Answer:

$Section = 4847.3205\ in^3$

Step-by-step explanation:

Given

$h = 15in$ -- height

$d = 22in$ -- diameter

Required

85% of the volume

First, calculate the radius

$r = \frac{1}{2} * d$

$r = \frac{1}{2} * 22= 11in$

The volume (V) is then calculated using:

$V = \pi r^2h$

$V = 3.142* 11^2 * 15$

$V = 5702.73\ in^3$

85% of the volume is:

$Section = 85\% * V$

$Section = 85\% * 5702.73\ in^3$

$Section = 4847.3205\ in^3$

3 0

2 years ago

Name the perpendicular lines

kotegsom [21]

Answer:

segment UV (put a line above it) and YW(put a line above it)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the equation is standard form, of the line passing through he points (2,-3) and (4,2)

Simora [160]

The point slope formula:

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

$(2,\ -3)\to x_1=2,\ y_1=-3\\(4,\ 2)\to x_2=4,\ y_2=2$

Substitute:

$m=\dfrac{2-(-3)}{4-2}=\dfrac{5}{2}=2.5\\\\y-(-3)=2.5(x-2)\\\\y+3=2.5x-5\ \ \ \ |+5\\\\y+8=2.5x\ \ \ \ \ \ |-y\\\\2.5x-y=8\ \ \ \ \ |\cdot2\\\\\boxed{5x-2y=16}$

5 0

3 years ago

Birds arrive at a birdfeeder according to a Poisson process at a rate of six per hour.

m_a_m_a [10]

Answer:

a) time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b) $P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c) $P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

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

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

$P(T>t)= e^{-\lambda t}$

a. What is the expected time you would have to wait to see ten birds arrive?

The original rate for the Poisson process is given by the problem "rate of six per hour" and on this case since we want the expected waiting time for 10 birds we have this:

time= $10 \frac{1}{6}=\frac{10}{6}=1.67 hours$

b. What is the probability that the elapsed time between the second and third birds exceeds fifteen minutes?

Assuming that the time between the arrival of two birds consecutive follows th exponential distribution and we need that this time exceeds fifteen minutes. If we convert the 15 minutes to hours we have 15(1/60)=0.25 hours. And we want to find this probability:

$P(T\geq 0.25h)$

And we can use the result obtained from the definitions and we have this:

$P(T\geq 0.25h)=e^{-(6)0.25}=0.22313$

c. If you have already waited five minutes for the first bird to arrive, what is the probability that the bird will arrive within the next five minutes?

First we need to convert the 5 minutes to hours and we got 5(1/60)=0.0833h. And on this case we want a conditional probability. And for this case is good to remember the "Markovian property of the Exponential distribution", given by :

$P(T \leq a +t |T>t)=P(T\leq a)$

Since we have a waiting time for the first bird of 5 min = 0.0833h and we want that the next bird will arrive within 5 minutes=0.0833h, we can express on this way the probability of interest:

$P(T\leq 0.0833+0.0833| T>0.0833)$

$P(T\leq 0.1667| T>0.0833)$

And using the Markovian property we have this:

$P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347$

3 0

3 years ago