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Masteriza [31]
3 years ago
13

The equation y-3=-2(x+5) is written in point-slope form. What is the y-intersept of the line?

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
8 0

Answer:

The correct answer would be B! -7

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20 POINTS help me with this question please
Leto [7]
<span>1.) Factor the denominator into 2 expressions leaves you with:

(x-2)/(x+5)(x-2)

x-2 cancels out, you are left with 1/(x+5), excluding x=2
</span>
7 0
3 years ago
Help me please?? ????
Mkey [24]

Answer:

40 days

Step-by-step explanation:

8 × 5 = 40

[character limit pass]

8 0
2 years ago
. One half of a number represented by x is -0.25. What is the value of x?
Jlenok [28]
Of means multiply
is means equals

1/2 • x = -.25

Rewritten

.5•x=-.25

Inverse operations to isolate the variable.

x = -.25/.5

x = -.5
7 0
3 years ago
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

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

8 0
2 years ago
If angle a has the terminal ray that falls in the fourth quadrant and cosine a equals 5/9 then determine the value of sin a in s
Tema [17]

Answer: -\dfrac{2\sqrt{14}}{9}

Step-by-step explanation:

Given

\cos a=\dfrac{5}{9}

a lies in the fourth quadrant

So, sine must be negative in the fourth quadrant

Using identity \sin ^2 x+\cos^2 x=1 to find sine value

\Rightarrow \sin^2 a=1-\dfrac{5^2}{9^2}

\\\\\Rightarrow \sin^2 a=1-\dfrac{25}{81}\\\\\\\Rightarrow \sin^2 a=\dfrac{56}{81}\\\\\\\Rightarrow \sin a=-\sqrt{\dfrac{56}{81}}\\\\\\\Rightarrow \sin a=-\dfrac{2\sqrt{14}}{9}

8 0
3 years ago
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