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erik [133]
2 years ago
8

This is the quest for the firs part of the other picture

Mathematics
2 answers:
Setler [38]2 years ago
7 0

Answer:

first one

Step-by-step explanation:

Mekhanik [1.2K]2 years ago
4 0
Abs are not a bad idea to heaven
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Find the midpoint of the segment.<br><br> A. (1.5, 2)<br> B. (6, 8)<br> C. (2, 1.5)
Ilia_Sergeevich [38]
The answer is A because it’s 1/5
7 0
3 years ago
Will Give brainliest!
Varvara68 [4.7K]

Answer:

It’s been a while.... I was send somewhere bad....

How are you

It’s chris Steven cummmings

And I’m sorry so sorry

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3 years ago
Solve the equation 3x + 4 = 5x - 8​
CaHeK987 [17]

Answer:

x = 6

Step-by-step explanation:

3x + 4 = 5x - 8

==> add 8 to both sides

in doing so we get 3x + 4 + 8 = 5x - 8 + 8

the -8 and the +8 cancels out and 4 + 8 = 12

we are left with 3x + 12 = 5x

==> subtract 3x from both sides

in doing so we get 3x - 3x + 12 = 5x - 3x

the 3x and -3x cancel out and 5x - 3x = 2x

we are left with 12 = 2x

==> divide both sides by 2

in doing so we get 12/2 which equals 6 and 2x/2x which leaves us with

6 = x

3 0
2 years ago
Read 2 more answers
Marla writes the problem, 628 divided by 1300.
Maslowich

Answer:

628/1300

Step-by-step explanation:

These are what would represent her problem..

8 0
3 years ago
Read 2 more answers
Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe
Stolb23 [73]

Answer:

Step-by-step explanation:

The model N (t), the number of planets found up to time t, as a Poisson process. So, the N (t) has distribution of Poison distribution with parameter (\lambda t)

a)

The mean for a month is, \lambda = \frac{1}{3} per month

E[N(t)]= \lambda t\\\\=\frac{1}{3}(24)\\\\=8

(Here. t = 24)

For Poisson process mean and variance are same,

Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8

 

(Poisson distribution mean and variance equal)

 

The standard deviation of the number of planets is,

\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828

b)

For the Poisson process the intervals between events(finding a new planet) have  independent  exponential  distribution with parameter \lambda. The  sum  of K of these  independent exponential has distribution Gamma (K, \lambda).

From the given information, k = 6 and \lambda =\frac{1}{3}

Calculate the expected value.

E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}\\\\=\frac{6}{\frac{1}{3}}\\\\=18

(Here, \alpha =k and \beta=\lambda)                                                                      

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г  (k)λ^k)(x)^(k-1).(e)^(-x/λ)

=\frac{1}{Г  (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%

Hence, the required probability is 0.2149 or 21.49%

5 0
3 years ago
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