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3 years ago

Coach Dan has decided to surprise the athletic

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

4 0

Havabauwvwishe i582283 ):&: 83?gegsuabe s

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Please help me i will give Brainly

Degger [83]

Answer:

Step-by-step explanation:

$\frac{3x + 5}{2x + 7}$ = 5

do cross multiplication

3x + 5 = 5(2x + 7)

3x + 5 = 10x + 35

5 - 35 = 10x - 3x

-30 = 7x

-30/7 = x

20 A

since there are 7 angles given it means that the polygon is heptagon as heptagon has 7 sides.

sum of interior angle of heptagon = (n-2)*180

(7-2)*180

5*180

900

Now ,

110 + 90 + 150 + 102 + 110 + 170 + x = 900

732 + x = 900

x = 900 - 732

x = 168 degree

20 B

since 5 angles are given it means that the polygon is pentagon as pentagon has 5 sides.

sum of interior angles of a pentagon = (n-2)*180

(5-2)*180

3*180

540 degree

Now ,

110 + 95 + 120 + 114 + x = 540

465 + x = 540

x = 540 - 465

x = 75 degree

let one rational number be x

according to the question,

1/7 * x = 2

x/7 = 2

do cross multiplication

x = 14

6 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

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

∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

8 0

2 years ago

18z 2+17z+4 I need to know what this is

balandron [24]

Answer:

18z 2+17z+4

Step-by-step explanation:

it hard try to give more point

4 0

3 years ago

What is -15÷3 please help me with this <br>

Sergio [31]

Answer:

-5

Step-by-step explanation:

You can work backwards with this problem if that helps you.

so, what times 3 = -15

-5 times 3 = -15 so that would be the answer

4 0

3 years ago

Read 2 more answers

Help me please I’m very confused

nadezda [96]

Answer:

$Arc \ length = 2\pi r\frac{\theta}{360} =2\pi 5\frac{30}{360} =10\pi \frac{1}{12} =\frac{5}{6} \pi =2.62units$

4 0

2 years ago