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3 years ago

F(x) = x4 - x3 - 2x- 4 factor the polynomial function over the complex numbers

Mathematics

1 answer:

olchik [2.2K]3 years ago

8 0

Answer:

Step-by-step explanation:

The given function is f(x) = x⁴ - x³ -2x -4

factor the polynomial function

f(x) = x⁴ - x³ -2x -4 = (x⁴ - 4) - (x³ + 2x ) ⇒ take (-) as a common from (- x³ -2x)

= (x² + 2 ) (x² - 2) - x (x² + 2) ⇒ take (x² + 2) as a common

= (x² + 2 ) ( x² - x - 2)

= (x + √2 i) (x - √2 i) (x -2 ) ( x+1)

Notes: (x⁴ - 4) = (x² + 2 ) (x² - 2)

(x² + 2)= (x + √2 i) (x - √2 i)

( x² - x - 2) = (x -2 ) ( x+1)

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Leave answer in the simplest radical form.

slega [8]

$\cfrac{7x^{-\frac{3}{2}} ~~ y^{\frac{5}{2}} ~~ z^{-\frac{2}{3}}}{56 x^{-\frac{1}{2}} ~~ y^{\frac{1}{4}}}\implies \cfrac{7}{56}\cdot \cfrac{y^{\frac{5}{2}} ~~ y^{-\frac{1}{4}}}{x^{-\frac{1}{2}} ~~ x^{\frac{3}{2}} ~~ z^{\frac{2}{3}}}\implies \cfrac{1}{8}\cfrac{y^{\frac{5}{2}-\frac{1}{4}}}{x^{-\frac{1}{2}+\frac{3}{2}}~~ z^{\frac{2}{3}}}$

$\cfrac{y^{\frac{9}{4}}}{8x z^{\frac{2}{3}}}\implies \cfrac{\sqrt[4]{y^9}}{8x\sqrt[3]{z^2}}\implies \cfrac{\sqrt[4]{y (y^2)^4}}{8x\sqrt[3]{z^2}}\implies \cfrac{y^2 \sqrt[4]{y}}{8x\sqrt[3]{z^2}}$

7 0

2 years ago

Please help 20 points. I will give Brainly to who ever get it right.

harkovskaia [24]

Answer:

Step-by-step explanation:

(-∞,2)

8 0

3 years ago

I’ll mark y brainlist for the correct answer!

inna [77]

Answer:

16 / C

Step-by-step explanation:

-4 x -4 = 16

4 x 4 = 16

two negatives cancel out when multiplying, that is why it equals 16*

5 0

3 years ago

Read 2 more answers

If the cost of carpeting a floor is $2.50 per square foot how much will it cost

Novosadov [1.4K]

Answer:

How many feet of the floor are you carpeting that is the real question

Step-by-step explanation:

It depends on how many feet you are putting the carpet on.

3 0

2 years ago