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2 years ago

7

(1+4i)−(−16+9i)

1 answer:

aniked [119]2 years ago

5 0

Answer:

there all cute i cant see the first one tho hella blurry she seems cute tho

Step-by-step explanation:

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An automobile insurance company issues a one-year policy with a deductible of 500. The probability is 0.8 that the insured autom

Effectus [21]

Answer:

y₀.₉₅ = 3659

Step-by-step explanation:

P( no accident ) = 0.8

P( one accident ) = 0

deductible = 500

mean = 3000

<u>Determine the 95th percentile of the insurance company payout </u>

Assuming : y =company payout , x =amount of loss incurred due to accident

Then :

P( x < 500 ) = 0.2 ( 1 - e^-500/3000)

= 0.2 ( 1 - e^-1/6 )

95th percentile =

= P( y < y₀.₉₅ ) 0.95

P( y = 0 ) = 0.8 + 0.2 ( 1 - e^-1/6 ) = 0.8307

attached below is the remainder of the solution

5 0

2 years ago

Use three fours to make 11.

mr Goodwill [35]

4 + 4 + 4 - 1 = 11? Was it supposed to be an equation? I am confused.

7 0

2 years ago

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

2 years ago

HELP PLEASE I NEED IT NOW!!!

Ostrovityanka [42]

Google them? google will have all of them

3 0

3 years ago

Solve for 2/5 + 3/8= With two possible answeres

Pani-rosa [81]

You can change the denominators so they are alike. $\frac{16}{40} + \frac{15}{40} = \frac{31}{40}$
Not sure how there can be two possible answers unless you're mentioning the decimal form:
.775
Tell me if this helps!!

6 0

2 years ago