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Len [333]
2 years ago
11

Which of the following situations could be described by the equation y = 120 - 25x?

Mathematics
1 answer:
Otrada [13]2 years ago
3 0

Answer:

Is there any other information?

Step-by-step explanation:

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Don’t understand this question will give brainless thing
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The answer is 6.33 units.
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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2
inna [77]
(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

6 x^{6}/3x = 2 x^{5} \\  \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\

We repeat previous steps until we run out of numbers:
3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2
\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )
6 0
3 years ago
Simplify 3(x-y)+3x+2<br>​
Alla [95]

Answer:

6x-3y+2

Step-by-step explanation:

I AM BIG BRAIN

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3 years ago
Suppose N has a geometric distribution with parameter p. Derive a closed-form expression for E(N | N &lt;= k), k = 1,2,... Check
vfiekz [6]

Answer:

P(X= k) = (1-p)^k-1.p

Step-by-step explanation:

Given that the number of trials is

N < = k, the geometric distribution gives the probability that there are k-1 trials that result in failure(F) before the success(S) at the kth trials.

Given p = success,

1 - p = failure

Hence the distribution is described as: Pr ( FFFF.....FS)

Pr(X= k) = (1-p)(1-p)(1-p)....(1-p)p

Pr((X=k) = (1 - p)^ (k-1) .p

Since N<=k

Pr (X =k) = p(1-p)^k-1, k= 1,2,...k

0, elsewhere

If the probability is defined for Y, the number of failure before a success

Pr (Y= k) = p(1-p)^y......k= 0,1,2,3

0, elsewhere.

Given p= 0.2, k= 3,

P(X= 3) =( 0.2) × (1 - 0.2)²

P(X=3) = 0.128

3 0
3 years ago
Can someone help me with my home work and add explanation if u can ill give up to 120. points if there right, this is one of the
Varvara68 [4.7K]

Answer:

(5,300)

Step-by-step explanation:

its the only one that its Y axis cant be divided by 5 and get a equal/even number

sorry I cant explain it well, but hope I somewhat helped ^^

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3 years ago
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