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tangare [24]
2 years ago
15

Angles A and B are supplementary. Angle A has a measure of 80°. What is the measure of angle B?

Mathematics
2 answers:
zheka24 [161]2 years ago
6 0

Answer:

100

Step-by-step explanation:

I just took the test

olganol [36]2 years ago
5 0
Supplementary angles: sum of the angles measure to be 180º
180º-A= B
Plug in...
180º-80º= 100º

So, angle B must be 100 degrees. 
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Triangle rst is dilated by a scale factor of 5 to form triangle r's't'. side r's' measures 40. what is the measure of side rs?
LenKa [72]

Answer: 8

Step-by-step explanation:

We know that scale factor = (image)/(preimage).

Substituting in what we have,

5=\frac{R'S'}{RS}\\\\5=\frac{40}{RS}\\\\RS=\boxed{8}

5 0
1 year ago
Increase 250ml by 85%
vampirchik [111]
The easiest way to do a problem like this is to think about the fact that you have 100% and you want to add another 85%.
This is 185% of the quantity that you have.
Multiply 250 by 185% or 1.85.
250 * 1.85 = 462.5 ml
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
Match the vocabulary word with its definition.
Alexxx [7]
1. Intersection
2. complement
3. Union
4. finite set
5. disjoint sets
6. proper subset
7. subset
7 0
3 years ago
A 15 foot flagpole casts an 11 foot shadow. At the exact same time a 28 foot tree casts a shadow. Which proportion would correct
marishachu [46]
The longer an object, the longer the shadow they cast at a given time of the day and vice versa. This is called direct proportionality. At the same time,

If 15 foot flagpole ---- casts 11 foot shandow
    28 foot tree -------- casts ?? foot shadow

The proportion would therefore be:

Length of shadow cast by tree = (28/15)*11 = 20.533 foot.
8 0
3 years ago
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