All categories

3 years ago

Use the graph of f shown in the figure below.

Mathematics

1 answer:

vaieri [72.5K]3 years ago

6 0

Answer:

Step-by-step explanation:

The attached graph shows the inverse function (red) and the value of the inverse function for an input of -9.

$f^{-1}(-9)=8$

_____

The inverse function tells you the input required for the given output from the function. So, you can look for an output of -9 on the graph of <em>f</em> and see what input gives that. You find ...

f(8) = -9

so ...

f⁻¹(-9) = 8

You might be interested in

9.5x - 2.5 for x = 9<br> What is the solution

Oksana_A [137]

Just substitue $9$ for $x$ in the expression.

$= 9.5(9) - 2.5$

$= 85.5 - 2.5$

$= 83$

7 0

2 years ago

Read 2 more answers

There are n machines in a factory, each of them has defective rate of 0.01. Some maintainers are hired to help machines working.

frosja888 [35]

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

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!

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!

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