What is the answer 9x plus 2 (2x+3)

Fynjy0 [20]

I really don’t know sorry I would help you but I can genesis hdbdbhs hebdbh

5 0

3 years ago

4. Automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul examines low-risk policies, continui

ANTONII [103]

Answer:

Step-by-step explanation:

for n ∈ N

Since Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping.

∴ the probability that Actuary Rahul examines exactly n policies

$(0.9)^{n-1}.(0.1)---(1)$

the probability that Actuary Toby examines more than exactly n policies

$(0.8)^n---(2)$

Given that policies are actually independent

∴ the probability that the event (1) and (2) happens simultaneously is

$(0.9)^{n-1}*(0.1)*(0.8)^n$

∴ the probability that Actuary Rahul examines fewer policies than Actuary Toby

$\sum ^\infty _{n=1} (0.9)^{n-1}*(0.1)*(0.8)^n\\\\=(\frac{0.1}{0.9} \sum ^\infty _{n=1}(0.72)^n\\\\=\frac{1}{9} (\frac{0.72}{0.28} )\\\\=\frac{2}{7} \\\\=0.2857$

<h3>the probability that Actuary Rahul examines fewer policies than Actuary Toby is 0.2857</h3>

5 0

3 years ago

A ship leaves port on a bearing of 34.0° and travels 10.4 miles. the ship then turns due east and travels 4.6 miles. how far is

maks197457 [2]

is a clockwise angle measured from due North. This is a problem, because all of the trigonometric functions are referenced to a counterclockwise angle measured from East.

A bearing of 34∘ corresponds to a trigonometric angle of θ1=90∘−34∘=56∘

The (x,y) values for the position of the ship after completing its first heading are:

x=(10.4mi)cos(56∘)
y=(10.4mi)sin(56∘)

The trigonometric angle for the second heading is θ2=90∘−90∘=0∘

The (x,y) values for the position of the ship after completing its second heading is:

x=(10.4mi)cos(56∘)+(4.6mi)cos(0∘)≈10.4mi
y=(10.4mi)sin(56∘)+(4.6mi)sin(0∘)≈8.6mi

The distance from port is:

d=√(10.4)2+(8.6)2≈13.5mi

Its trigonometric angle is:

θ=tan−1(yx)

θ=tan−1(8.610.4)

θ≈39.6∘

The bearing angle is:

θb=90∘−39.6∘=50.4∘

6 0

4 years ago

Explain why each of the following integrals is improper. (a) 4 x x − 3 dx 3 Since the integral has an infinite interval of integ

erma4kov [3.2K]

Answer:

a

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

b

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

c

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

d

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Step-by-step explanation:

Considering a

$\int\limits^4_3 {\frac{x}{x- 3} } \, dx$

Looking at this we that at x = 3 this integral will be infinitely discontinuous

Considering b

$\int\limits^{\infty}_0 {\frac{1}{1 + x^3} } \, dx$

Looking at this integral we see that the interval is between $0 \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering c

$\int\limits^{\infty}_{- \infty} {x^2 e^{-x^2}} \, dx$

Looking at this integral we see that the interval is between $-\infty \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering d

$\int\limits^{\frac{\pi}{4} }_0 {cot (x)} \, dx$

Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral

7 0

3 years ago

PLEASE Help Fast I’m Struggling!

meriva

Answer:

y=-2

x=-6

Step-by-step explanation:

4 0

3 years ago