Please help me I really need it

What is the modulus after (-1+sqrt3i) gets raised to the 6th power

Eddi Din [679]

Answer:

it is b

Step-by-step explanation:

took the test

8 0

2 years ago

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Are these parallelograms congruent? 4 cm 6 cm 6 cm 9 cm yes no

lianna [129]

Answer:

Step-by-step explanation:

Yes 6 over 9 simplifys to 2 over 3.

And 4 over 6 simplifys down to 2 over 3

Can I have brainliest please :D

7 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

Which polygon will not tessellate a plane?

netineya [11]

Answer:

Regular Octagon

Step-by-step explanation:

There are only three regular shapes that tessellate – the square, the equilateral triangle, and the regular hexagon. All other regular shapes, like the regular pentagon and regular octagon, do not tessellate on their own.

7 0

2 years ago

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Jamal earns $40 washing 5 windows. At this rate, how many windows did Jamal wash to earn $80?

Alja [10]

10 windows : divide 40/5=8
Then divide 80/8 =10

3 0

2 years ago

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