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3 years ago

(7i)(7i)help! please show steps

Mathematics

1 answer:

exis [7]3 years ago

7 0

Answer:

-49

Step-by-step explanation:

multiply 7 with 7 then i with i and lastly multiply the result numbers

7i × 7i ➡ 7 × 7 = 49

i × i = i^2

49 × i^2 = 49i^2

i^2 = -1 ➡ 49 × (-1) = -49

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Answer:

a) 7

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Step-by-step explanation:

Just put the opposite for each one. The absolute value is the distance away from 0.

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2 years ago

-5=-2/9x<br>I dont know the answer

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-5=-2/9x.

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3 years ago

Read 2 more answers

Ten percent of U.S. households contain 5 or more people. You want to simulate choosing a household at random and recording "Yes"

Debora [2.8K]

Answer:

(b) 0 = Yes; 1-9 = No

Step-by-step explanation:

Among digits, 10% of 10 digits is a single digit. Any answer choice that suggests more than one digit be used for "yes" is not appropriate. A good choice is ...

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3 years ago

Which expression is equivalent to<br> -32^3/5

Tanya [424]

Answer:

Step-by-step explanation:

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2 years ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

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2 years ago

(7i)(7i)help! please show steps​

(7i)(7i)help! please show steps