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2 years ago

Find the root of 5x - 3 = 0

Mathematics

1 answer:

larisa86 [58]2 years ago

7 0

The root of 5x-3=0 is (3/5,0)

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PLEASE HELP!!

SIZIF [17.4K]

(x, y)

Domain = x

Range = y

It is a function when none of the x values are repeating.

Answer - C) Both A and B

Hope this helps and makes sense!!

8 0

2 years ago

Seven-ninths of the pencils in a box are yellow. Three-tenths of the yellow pencils are sharpened. What fraction represents the

jek_recluse [69]

Answer:

7/30

Step-by-step explanation:

We are given that Seven-ninths (7/9) of the pencils in a box are yellow and three-tenths (3/10) of the yellow pencils are sharpened.

To find the fraction that represents sharpened yellow pencils, we find the product of the the fraction of yellow pencils and the fraction of yellow pencils that are sharpened.

That is:

=> (7 / 9) * (3 / 10)

=(7 * 3) / (9 * 10)

= 21 / 90

= 7 / 30

7/30 represents the fraction of sharpened yellow pencils.

3 0

3 years ago

What is the area of the square?

NISA [10]

The area is 4 square meters, because 2x2 would be 4

7 0

2 years ago

Help!!!!!! I’ve been working on this for 1 hour

kifflom [539]

Answer:

7.83

Step-by-step explanation:

cos(40)=6/(ZX)

zx=7.83

5 0

3 years ago

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of si

OverLord2011 [107]

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

$P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!$

a) The probability that exactly 3 arrivals occur during a particular hour is:

$P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\$

b) The probability that <em>at least</em> 3 people arrive during a particular hour is:

$P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938$

c) In this case, t=0.25, so we recalculate the parameter as:

$\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5$

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

$E(x)=\lambda=1.5$

3 0

2 years ago