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ahrayia [7]
3 years ago
10

15x − 2 =? please help

Mathematics
1 answer:
pentagon [3]3 years ago
8 0

Answer:

I cant tell is the 'x' is supposed to be a variable or a multiplication sign. IF its a multiplication sign then the answer is -30. Btu if its a variable then its in its simplest form there isn't any way to solve that equation.

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When multiplying two polynomials of degrees m and n, will the product be a polynomial? If so, explain, and state the degree.
vfiekz [6]
Yes.
The degree will be m+n
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3 years ago
Identify the radicand in the radical below
Elena L [17]

Answer:

The radicand is whatever is under the the radical, so in this case, your answer would be D. 3x-5



4 0
3 years ago
Which comparison is not correct?
Elden [556K]

Answer:

it’s the third one

Step-by-step explanation:negatives are never higher than numbers without negatives

7 0
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galina1969 [7]
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4 0
2 years ago
The median of a probability distribution can be defined as the number m such that Upper P (Upper X less than or equals m )equals
Alexxandr [17]

Answer:

tex]M=\beta ln(2)[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate).

Solution to the problem

For this case we can use the following Theorem:

"If X is a continuos random variable of the exponential distribution with parameter \beta for some \beta \in R >0"

Then the median of X is \beta ln (2)

Proof

Let M the median for the random variable X.

From the definition for the exponential distribution we know the denisty function of X is given by:

f_X (x) = \frac{1}{\beta} e^{-\frac{x}{\beta}}

Since we need the median we can put this equation:

P(X

If we evaluate the integral we got this:

\frac{1}{\beta} \int_0^M e^{- \frac{x}{\beta}}dx =\frac{1}{\beta} [-\beta e^{-\frac{x}{\beta}}] \Big|_0^M

And that's equal to:

1/2 = 1 -e^{- \frac{M}{\beta}}

And if we solve for M we got:

1-e^{- \frac{M}{\beta}} = \frac{1}{2}

e^{- \frac{M}{\beta}}=\frac{1}{2}

If we apply natural log on both sides we got:

-\frac{M}{\beta}=ln(1/2)

And then M=\beta ln(2)

5 0
3 years ago
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