Is 1/2 closest to 0 or 1

(HELP! PLEASE!!! 50 POINTS!! JUST TWO QUESTIONS) 1. Using the numbers 5, 8, and 24, create a problem using no more than four ope

andreev551 [17]

Answer:Davinki

Step-by-step explanation:made the mona lisa

4 0

3 years ago

-6x + 14 < -28<br> AND 3x + 28 < 25

trasher [3.6K]

Answer:

x>7

-6x+14< -28

-14. -14

-6x<-42

÷-6 ÷-6

x<7

switch signs because we divided by a negative number

x>7

Answer: x<-1

3x+28<25

-28 -28

3x<-3

÷3 ÷3

x<-1

7 0

3 years ago

Mrs.Cheng has $18 deducted monthly from checking account for her gym membership. What integer represents the change in her accou

Vikki [24]

First you would have to multiply $18 (The cost for each month) and 12 (the number of months in a year) and the answer would be 216
So she deducts 216 a year, so the answer would have to be negative

The final answer is -216

4 0

3 years ago

What is w + 5? I need help!

alukav5142 [94]

5w if it did not give a problem where you could figure out what w is.

8 0

3 years ago

Read 2 more answers

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h

saveliy_v [14]

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

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

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\$

The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is

$EV=\lambda*t=5*0.75=3.75$

8 0

3 years ago