If a small bag of Caramel Bliss weighs 12 ounces and an 8 week old kitten weighs 1.5 pounds, then how many kittens does it take
1 answer:
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Z
Z=kX
When x=6, z=18,
Finding value of z,
18=k*6
k=3
Z=3X
when x=10,
Z=3*10
Z=30
Part B;
Z
z=
Where x is a constant of proportionality,
when x=3, z=2
2=
a=3*2
a=6
When x is 13,
Z=
Z=kX
When x=6, z=18,
Finding value of z,
18=k*6
k=3
Z=3X
when x=10,
Z=3*10
Z=30
Part B;
Z
z=
Where x is a constant of proportionality,
when x=3, z=2
2=
a=3*2
a=6
When x is 13,
Z=
Answer:
a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
Step-by-step explanation:
Given that;
if n ⇒ ∞
p ⇒ 0
⇒ np = Constant = λ, we can apply poisson approximation
⇒ Here 'p' is small ( p=0.01)
⇒ if (n=large) we can approximate it as prior distribution
⇒ let the number of defective items be d
so p(d) = ((e^-λ) × λ) / d!
NOW
a)
Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.
λ[n0.01]
p[ d > 20x ] = 1 - [ d ≤ 20x ]
= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b)
Similarly in this case if number of machines d > 80x/3;
Then it can not be repaired in time
p[ d > 80x/3 ]
1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c)
n = 300, lets do it for first case i.e;
p [ d > 20x } ≤ 0.01
1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01
⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99
⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ
∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
<em>Answer:</em>
<em>$0.38 per yard</em>
<em>Step-by-step explanation:</em>
<em>Step 1:</em>
Given; A 9-yard roll of string that costs $3.42.
Required; To find the unit price
<em> Step 2:</em>
Convert $3.42 to cents
<em>Step 3:</em>
<em>Find the unit price</em>
<em>Convert to dollars</em>
Hence the unit price is $0.38 per yard.