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3 years ago

Estimate 38 3/22 - 14 29/30 by rounding each number to the nearest whole number

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

38-15= 23. I’m guessing this is what you meant. If not please correct me.

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Plz help me!!!!!!!!!!!!!

tigry1 [53]

If the formula is a+b-c then you need to substitute the letters with the numbers. For this equation the formula would be:
(1+3-4) + (2+5-6)
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The answer would be 1

4 0

2 years ago

Read 2 more answers

Hey guys I need help

Gemiola [76]

Answer:

20.8

Step-by-step explanation:

7 0

2 years ago

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Furkat [3]

Imagine a ladder leaning against the wall and you know the height of the ladder and its distance from the wall and you had to find out the height you could use pythagorean theorem here

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3 years ago

Read 2 more answers

Brainliest to first answer correct.

suter [353]

Answer:

Option A is correct.

Step-by-step explanation:

The height of the rectangular pack is h1 = 4r

The height of the parallelogram = 2r + the height of equilateral triangles with a side of 2r

The height of equilateral triangle with a side of 2r = (side)/2*√3

=(2r)/2 *√3 = r√3

The height of the parallelogram h2 = 2r+r√3 = r(2+√3)

So,the ratio of h1 to h2 is :

4r : r(2 +√3)

Cancelling r we get

Hence, the ratio is 4 : 2 +√3

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

3 years ago