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monitta
3 years ago
5

The degree of the polynomial function below is ____ f(x)=-3x^5+4x-2

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0

Answer:

the degree of the polynomial is 5

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A geologist in South America discovers a
diamong [38]

Answer:

This is a radioactive decay / half-life problem.

Initial amount of C 14 = 100% Present Amount = 57%

k = .0001  (that value should be negative)

Nt = No * e^ (k*t)

You need to solve that equation for "time" (equation attached)

time = natural log (Ending amount / Starting Amount) / k

time = natural log (57% / 100%) / -.0001

time = ln (.57) / -.0001

time = -.56211891815 / -.0001

time = 5,621.2 years    (age of the bird skeleton)

These problems are quite complicated but I think I know this pretty well.

Need to know more? Visit my website (it's on the graphic).

Step-by-step explanation:

7 0
3 years ago
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
3 years ago
(Will give brainliest if asked for)
GarryVolchara [31]

A. x <= -12

<u>Explanation:</u>

-3x >= 36

-x >= 36/ 3

-x > = 12

x < = -12

When the negative sign is shifted then the greater than equal to sign reverses to less than equal to.

Therefore, x <= -12

7 0
3 years ago
The fish population in a local stream is decreasing at a rate of 3% per year. The original population was 48,000. Write an expon
ElenaW [278]

Answer:

Results are below.

Step-by-step explanation:

Giving the following information:

Decrease rate (d)= 7%

Number of periods (n)= 7 years

Current population (PV)= 48,000

<u>First, to calculate the future value, we need to use the following decrease exponential formula:</u>

<u />

FV= PV*[(1+d)^-n]

<u>After 7 years:</u>

FV= 48,000*(1.07^-7)

FV= 29,892

8 0
2 years ago
Which of the following values are solutions to the<br> inequality –8 + 2x &gt; 5
myrzilka [38]

Answer:

6

Step-by-step explanation:

i guess

3 0
2 years ago
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