X/3=9 X=??? I need help!

wlad13 [49]

It's a linear function with a slope 0.

It's a horizontal line (look at the picture).

4 0

3 years ago

Igoryamba

I : 2n + 1
II : 2n + 3
III : 2n +5
(2n + 1) + (2n + 3) +(2n+5)=63
6n+9=63
n=9
I : 2n+1=9*2+1=19
II : 2n+3=9*2+3=21
III : 2n+5=9*2+5=23

19+21+23=63

5 0

3 years ago

lys-0071 [83]

He would need 10 carrots and 5 potatoes for 15 servings.

Hope this helps.

7 0

4 years ago

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

Westkost [7]

He should measure in liters
Hope it is correct

6 0

3 years ago