Find the solution set.<br> x? + 20x + 100 = 0<br> Enter the correct answer.

Write -1/9 + 10x - 2x^3 + x^5 - x ^4 in standard form

matrenka [14]

X^5 - x^4 - 2x^3 + 10x - 1/9 is the standard form

5 0

3 years ago

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What expression is not equivalent to 2/3 *4.

Gelneren [198K]

Answer:

Step-by-step explanation:

Any expression that says (2 * 4)/3 is correct, even though there is a shift in the brackets. That means that A, B, C are all the same thing, and all are correct.

Only D is lying.

2*(1/3) = 2/3

4*(1/3) = 4/3 Now we have to add them

2/3 + 4/3 = 6/3 = 2

To show what the real answer is, try A

2*4/3

8/3 is the actual answer.

4 0

3 years ago

Pls we only need part B to pass this subject

ikadub [295]

Answer: Weird

Step-by-step explanation: Maybe calculate all the numbers I guess.

8 0

3 years ago

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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

(c) 90.53%

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 $p_2$ .