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Igoryamba
3 years ago
13

HEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
What is the coefficient of the fourth term in this expression?
m3+2kn2+mn2+6k2n
Mathematics
1 answer:
lesantik [10]3 years ago
5 0
The coefficients would be 6 and 2.
Coefficient of k would be 6 and for n it would be 2.
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A lab animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial,
Tresset [83]

Answer:

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

Step-by-step explanation:

Given - A lab animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial, it will choose the same food on the next trial with a probability of 50%, and it will choose the other foods on the next trial with equal probabilities of 25%.

To find - If the animal chooses food #1 on an initial trial, what is the probability that it will choose food #2 on the second trial after the initial trial?

Proof -

By the given information, we get the stohastic matrix

H = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]

As we know that,

The matrix is a Markov chain x_{k+1} = Hx_{k}

Let

The initial state vector be

x_{0} = \left[\begin{array}{ccc}1\\0\\0\end{array}\right]

we choose this initial vector because given that If the animal chooses food #1 on an initial trial.

Now,

x_{1} = Hx_{0} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}1\\0\\0\end{array}\right] \\= \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

∴ we get

x_{1} = \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

Now,

x_{2} = Hx_{1} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right] \\= \left[\begin{array}{ccc}0.25+0.0625+0.0625\\0.125+0.125+0.0625\\0.125+0.0625+0.125\end{array}\right]\\= \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

x_{2} = \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

4 0
3 years ago
(02.03)Which statement is true about the result of a rigid transformation?
oksian1 [2.3K]

Answer:

the preimage will be congruent to the image

Step-by-step explanation:

4 0
3 years ago
Jonas spends an average of $3.42 on meals each day. How much does he spend on meals each week?
LenaWriter [7]
He Spends $24.64 Every Week
Hope This Helps
4 0
4 years ago
Read 2 more answers
A sequence is defined by the recursive formula f (n 1) = f(n) – 2. if f(1) = 18, what is f(5)?
Marat540 [252]
Would assume:

f(n + 1) = f(n) - 2

f(2) = f(1 + 1) = f(1) - 2 = 18 - 2 = 16

<span>f(3) = f(2 + 1) = f(2) - 2 = 16 - 2 = 14
</span>
<span>f(4) = f(3 + 1) = f(3) - 2 = 14 - 2 = 12
</span>
<span>f(5) = f(4 + 1) = f(4) - 2 = 12 - 2 = 10
</span>
f(5) = 10
4 0
3 years ago
PLZ HELP!!!
dexar [7]

Solution:

The given equation is:

3^2 x + 7^2= 2 ^x\\\\9 x +49 =2^x

A: Taking , x=1

L H S = 9 × 1  + 49 = 9 + 49= 58

R H S : 2^x=2^1=2

L H S ≠   R H S

B: Taking x=3

L H S = 9 × 3 + 49=27 + 49=76

R HS = 2^x = 2^3 = 8

L H S ≠  R H S

C: Taking x=5

L H S = 9 × 5 + 49

        = 45 + 49

       = 94

R H S:

2^x=2^5=32

L H S ≠ R H S

(D)

when , x= 8

L H S

9 × 8 + 49

= 72 + 49

= 121

R HS:

2^x=2^8 =256

L H S ≠ R H S






8 0
3 years ago
Read 2 more answers
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