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3 years ago

Determine if the expression – 6y 25 – is a polynomial or not. If it is a

Mathematics

1 answer:

harkovskaia [24]3 years ago

6 0

Answer:

7346

Step-by-step explanation:

hopefuly this help you!

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X=y+4 and y=x+4

timofeeve [1]

Alright, for the two equations you gave me, which are x=y+4 and y=x+4, we can try substitution to get the answer! Substitution is when you take one variable from one equation and plug it into another equation to find out what both variables are!

To do this, we start with the first equation, x=y+4. Then, from the second equation we know that y=x+4 so we can plug x+4 in for the y in the first equation. We now have x=(x+4)+4 as our first equation.

Now, we can simplify! The parenthesis aren't important in this equation, so we can just get rid of them, giving us x=x+4+4. Now we can simplify to give us x=x+8. Now, all that's left is to subtract the x from both sides of the equation! This gives us 0=8, but we know that that's not possible because there is no way that 0 can equal 8. This means that this equation doesn't have an answer!

To check this, you can repeat the steps above with the second equation, but you will still end up with 0=8, and that, again, means that this equation doesn't have an answer.

6 0

4 years ago

Read 2 more answers

Erica bought a car for $24,000. She had to add Pennsylvania’s sales tax of 6%. The total price of the car is closest to?

Arlecino [84]

I believe it may be c

6 0

4 years ago

Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

a. Then, for $\hat \theta_3$ to be an unbiased estimator ; Then:

$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

Thus; in order to minimize the variance of $\hat \theta_3$ ; then constant a can be determined as :

$V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2$

Using differentiation:

$\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0$

⇒

$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

$\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}$

This implies that

$\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0$

So, $V( \hat \theta_3)$ is minimum when $\hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}$

As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

4 0

4 years ago

Find the slope of the line that goes through the points (-1,-4) and (9, 20). You must show all of your work in the space provide

Setler [38]

Answer:

12/5

Explanation:

To find the slope of the line, we'll need to use the below formula;

$m=\frac{y_2-y_1}{x_2-x_1}$

where x1 = -1, y1 = -4, x2 = 9, y2 = 20, so let's go ahead and substitute these values into our formula and solve for m;

$m=\frac{20-(-4)}{9-(-1)}=\frac{20+4}{9+1}=\frac{24}{10}=\frac{12}{5}$

5 0

1 year ago

A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess

statuscvo [17]

Let's start by visualising this concept.

Number of grains on square:
1 2 4 8 16 ...

We can see that it starts to form a geometric sequence, with the common ratio being 2.

For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
$T_n = ar^{n - 1}$
$T_{15} = 2^{14} = 16384$

Thus, there are 16, 384 grains on the fifteenth square.

The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
$S_n = \frac{a(r^{n} - 1)}{r - 1}$
$S_{15} = \frac{2^{15} - 1}{2 - 1}$
$S_{15} = 2^{15} - 1 = 32767$

Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.

6 0

3 years ago

Read 2 more answers