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asambeis [7]
3 years ago
9

Use the distributive property to simplify the expressions. 1B. -3(9x-q) 2B. 2(12+5p) 3b. -7(-8y+7) 4b. 10(2+7c) 5b. -8(7+11k) 6b

. -(3 - 7u) 7b. -6(5p + s)
Mathematics
1 answer:
Debora [2.8K]3 years ago
8 0
1B: -3(9x-q)
       -27x + 3q


Step 1: Distribute the -3 to the 9x to get -27x.
Step 2: Distribute the -3 to the -q.


2B: 2(12+5p)
       24 + 10p

Step 1: Distribute the 2 to the 12 to get 24.
Step 2: Distribute the 2 to the 5p to get 10p.

3B: -7(-8y+7)
       56y - 49


Step 1: Distribute the -7 to the -8y to get 56y.
Step 2: Distribute the -7 to the 7 to get -49.

4B: 10(2+7c)
        20 + 70c


Step 1: Distribute the 10 to the 2 to get 20. 
Step 2: Distribute the 10 to the 7c to get 70c.

5B: -8(7+11k)
       -56 - 88k


Step 1: Distribute the -8 to the 7 to get -56.
Step 2: Distribute the -8 to the 11k to get -88k.

6B: -(3-7u)
       -1(3-7u)
       -3 + 7u


Step 1: Place a 1 after the negative symbol to symbolize -1.
Step 2: Distribute the -1 to the three to get -3.
Step 3: Distribute the -1 to the -7u to get 7u.


7B: -6(5p + s)
       -30p - 6s

Step 1: Distribute the -6 to the 5p to get -30p.
Step 2: Distribute the -6 to the s to get -6s.

:) :D




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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)

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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%

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rounding it to the nearest degree will also be 20

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