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4 years ago

What is a polynomial function in standard form with zeroes 1, 2, -2. and -3

1 answer:

3 0

<span>A polynomial with the given zeros can be represented as
f(x) = (x-1)(x-2)(x+2)(x+3).
Note that if you set f(x) = 0, then 1,2,-2, and -3 certainly are the solutions. From here, we simply multiply/expand out the polynomial. We can do this in a variety of ways, one of which is taking the left two and right two products separately. We have
(x-1)(x-2) = x^2 - 3x + 2
and
(x+2)(x+3) = x^2 + 5x + 6.
This gives that
f(x) = (x^2 - 3x + 2) (x^2 + 5x + 6).
Expanding this expression out then gives us our answer as
f(x) = x^4 + 2x^3 - 7x^2 - 8x + 12
or answer choice B.</span>

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Will give brainlest answer pls help!

mojhsa [17]

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4 years ago

Please help I will give brainliest!

Burka [1]

Answer: the statement “measure of <4 is 105 degrees” is incorrect.

Explanation: This is because since both angles 1 and 3 are vertical, this means they are both 95. This gives us 190 (95x2). Now we subtract this from 360 which gives us 170. Since angles 2 and 4 are vertical, we divide by 2, which gives us 85, not 105.

Hope this helps! :)

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3 years ago

Is 64^2−48+9 a perfect square trinomial?

Marysya12 [62]

Answer:

yes, it is a perfect square trinomial because the square of binomial are (x-0.375) (x-0.375)

5 0

3 years ago

Solve for x : 7x-8a = 3x+24a

Bingel [31]

Answer:

4x=32a

Step-by-step explanation:

7 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

4 years ago