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yarga [219]
3 years ago
9

11 + 11 = 4 22 + 22 = 16 33 + 33 = ?

Mathematics
2 answers:
Kryger [21]3 years ago
7 0

Answer:

Answer will be 36

Step-by-step explanation:

11*11=4 -> (1+1)*(1+1)=4, 22*22=16 -> (2+2)*(2+2)=16. Hence, 33*33 -> (3+3)*(3+3)=36. Add the digits of 11 that will be 2. hope this helps you :)

Anon25 [30]3 years ago
4 0

Answer:

66

Step-by-step explanation:

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What is the mean of -71,-62,-88-38-89,15,-76
olga nikolaevna [1]
71
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88
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6 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}

<em>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. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

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

5 0
3 years ago
The radius of a circle is 12 in. Find its area in terms of \piπ.
iren [92.7K]

Answer:

pi144

Step-by-step explanation:

square the radius but dont multiply by pi because the question wants you to leave it out.

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3 years ago
What’s the answer pls help me
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If the function g has the factors (x-7) and (x+6), what are the zeros of function g?
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The zeros are (7,0) and (-6,0).
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