All categories

3 years ago

Choose the expression that completes the equation.

Mathematics

2 answers:

In-s [12.5K]3 years ago

6 0

Answer:

Option 3

Step-by-step explanation:

33 / 3 + 10 = 21

1.) 5(3 + 5) = 40

2.) 10(1 + 1) = 22

3.) 3(6 + 1) = 21

Harrizon [31]3 years ago

5 0

Answer:

Answer would be 3

Step-by-step explanation:

2(10+1)= 21 and

33 divided by 3 + 10 = 21

You might be interested in

What is the slope line of m (0,6),(3,0)

mezya [45]

Answer:

slope= y²-y¹

x²-x¹

0-6

3-0

-6

= -2

3 0

3 years ago

Ok, so I gotta find the area for these rooms, but its not clear where the rooms end.. Where would you guys say the living room e

notsponge [240]

Measure your foot size then go to the wall barefooted and walk in a straight line to the door the wall to wall the against the wall then add

8 0

3 years ago

Read 2 more answers

What is the unit rate for the ratio 500 apples/4 bushels?

vovikov84 [41]

C, I’m pretty sure

Why do I have to write 20 words for this? This is why I’m rambling.

7 0

3 years ago

If there are 22 red cards and 28 black card, what is the probability you would draw a red card?

morpeh [17]

it would be 22/50 or 11/25

3 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 at least 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